Case Study / Scenario-Based MCQs for Sub-Topics of Topic 12: Vectors & Three-Dimensional Geometry

Question 1. A person walks 3 km East and then 4 km North. Which of the following statements is/are true regarding their journey?

Scenario: Let East be along the positive x-axis ($\hat{i}$) and North be along the positive y-axis ($\hat{j}$). The displacement vectors are $\vec{d}_1 = 3\hat{i}$ km and $\vec{d}_2 = 4\hat{j}$ km.

(A) The total distance covered is 7 km.

(B) The net displacement is a vector $7\hat{i}\hat{j}$ km.

(C) The magnitude of the net displacement is 5 km.

(D) The direction of the net displacement is towards the North-East.

Question 2. Two forces $\vec{F}_1 = 2\hat{i} + 3\hat{j}$ N and $\vec{F}_2 = 3\hat{i} - \hat{j}$ N act on an object. What is the resultant force acting on the object?

Scenario: The resultant force is the vector sum of the individual forces.

(A) $5\hat{i} + 2\hat{j}$ N

(B) $\hat{i} + 4\hat{j}$ N

(C) A scalar quantity.

(D) A vector with magnitude $\sqrt{29}$ N.

Question 3. A boat travels with a velocity of $\vec{v}_b = 4\hat{i}$ km/h in still water. The river flows with a velocity of $\vec{v}_r = 3\hat{j}$ km/h. What is the resultant velocity of the boat relative to the ground?

Scenario: The resultant velocity is the vector sum of the boat's velocity relative to water and the river's velocity.

(A) $7$ km/h

(B) $4\hat{i} + 3\hat{j}$ km/h

(C) $5$ km/h

(D) A vector pointing North-East.

Question 4. A vector $\vec{a}$ has magnitude 10. A new vector $\vec{b}$ is obtained by multiplying $\vec{a}$ by a scalar $-2$. What is the magnitude and direction of $\vec{b}$?

Scenario: Scalar multiplication scales the magnitude by the absolute value of the scalar and reverses direction if the scalar is negative.

(A) Magnitude is $-20$.

(B) Magnitude is 20.

(C) Direction is the same as $\vec{a}$.

(D) Direction is opposite to $\vec{a}$.

Question 5. A vector $\vec{p}$ represents a displacement of 5 units towards East. A vector $\vec{q}$ represents a displacement of 5 units towards West. Which statement describes the relationship between $\vec{p}$ and $\vec{q}$?

Scenario: Consider East as positive and West as negative along the x-axis.

(A) They are equal vectors.

(B) They are collinear vectors.

(C) $\vec{q} = -\vec{p}$.

(D) $|\vec{p}| = |\vec{q}|$.

Question 6. Three vectors $\vec{A}, \vec{B}, \vec{C}$ form the sides of a closed triangle, taken in order (head-to-tail). What is the sum of these three vectors?

Scenario: Apply the triangle law of vector addition sequentially.

(A) $\vec{A} + \vec{B} + \vec{C}$ has the same magnitude as $\vec{A}$.

(B) The resultant vector is the zero vector $\vec{0}$.

(C) The magnitude of the sum is zero.

(D) The sum points from the start of $\vec{A}$ to the end of $\vec{C}$.

Question 7. An object is pulled by two ropes. Rope 1 applies a force of 100 N due East. Rope 2 applies a force of 100 N due North. What is the magnitude of the total force on the object?

Scenario: Forces are vectors. The total force is the vector sum. Use components if needed.

(A) 200 N

(B) 0 N

(C) $\sqrt{100^2 + 100^2} = \sqrt{20000} = 100\sqrt{2}$ N

(D) $100\sqrt{2}$ N

Question 1. Point P has coordinates (2, 3) in a 2D Cartesian system. What is the position vector of P relative to the origin?

Scenario: The position vector of a point $(x,y)$ is $x\hat{i} + y\hat{j}$.

(A) $2\hat{i} + 3\hat{j}$

(B) $5$

(C) $(2, 3)$

(D) $\sqrt{13}$

Question 2. A vector starts at point A (1, 2, 0) and ends at point B (3, -1, 4). Find the components of the vector $\vec{AB}$.

Scenario: The components of $\vec{AB}$ are the differences in coordinates, $(x_B-x_A, y_B-y_A, z_B-z_A)$.

$\vec{AB} = (3-1)\hat{i} + (-1-2)\hat{j} + (4-0)\hat{k} = 2\hat{i} - 3\hat{j} + 4\hat{k}$.

(A) $(2, -3, 4)$

(B) $2\hat{i} - 3\hat{j} + 4\hat{k}$

(C) $\sqrt{4+9+16} = \sqrt{29}$

(D) $(-2, 3, -4)$

Question 3. Determine if the points A(1, -2, 3), B(2, 1, -4), and C(0, -5, 10) are collinear using vector components.

Scenario: Three points A, B, C are collinear if $\vec{AB}$ is a scalar multiple of $\vec{AC}$. Calculate $\vec{AB}$ and $\vec{AC}$ and check proportionality of components.

$\vec{AB} = (2-1)\hat{i} + (1-(-2))\hat{j} + (-4-3)\hat{k} = \hat{i} + 3\hat{j} - 7\hat{k}$.

$\vec{AC} = (0-1)\hat{i} + (-5-(-2))\hat{j} + (10-3)\hat{k} = -\hat{i} - 3\hat{j} + 7\hat{k}$.

The components are $(1, 3, -7)$ for $\vec{AB}$ and $(-1, -3, 7)$ for $\vec{AC}$. The ratios are $\frac{-1}{1} = -1$, $\frac{-3}{3} = -1$, $\frac{7}{-7} = -1$. Since the ratios are equal, the vectors are collinear, and thus the points are collinear.

(A) Yes, they are collinear because $\vec{AC} = -\vec{AB}$.

(B) No, they are not collinear because their components are different.

(C) Yes, because the ratio of corresponding components is constant ($1/-1 = 3/-3 = -7/7 = -1$).

(D) Yes, because they satisfy the equation of a single line.

Question 4. Express the vector $\vec{v} = 7\hat{i} - \hat{j}$ as a linear combination of $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - \hat{j}$. Find the scalar coefficients $x$ and $y$ such that $\vec{v} = x\vec{a} + y\vec{b}$.

Scenario: Substitute the component forms and equate coefficients of $\hat{i}$ and $\hat{j}$ to form a system of linear equations.

$7\hat{i} - \hat{j} = x(\hat{i} + \hat{j}) + y(\hat{i} - \hat{j}) = (x+y)\hat{i} + (x-y)\hat{j}$.

Equations: $x+y = 7$ and $x-y = -1$. Adding them gives $2x=6 \Rightarrow x=3$. Substituting $x=3$ into the first equation gives $3+y=7 \Rightarrow y=4$.

(A) $x=3, y=4$

(B) $x=4, y=3$

(C) $\vec{v} = 3(\hat{i} + \hat{j}) + 4(\hat{i} - \hat{j})$

(D) This is possible because $\vec{a}$ and $\vec{b}$ are non-collinear vectors in 2D.

Question 5. Find the magnitude of the vector $\vec{AB}$ where A is $(1, -1, 2)$ and B is $(2, 1, -1)$.

Scenario: First find the vector $\vec{AB}$ in component form, then calculate its magnitude using the formula $\sqrt{x^2+y^2+z^2}$.

$\vec{AB} = (2-1)\hat{i} + (1-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + 2\hat{j} - 3\hat{k}$.

$|\vec{AB}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1+4+9} = \sqrt{14}$.

(A) $\sqrt{11}$

(B) $\sqrt{14}$

(C) $\sqrt{17}$

(D) $14$

Question 6. Three vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j} + \hat{k}$, and $\vec{c} = 2\hat{i} + 3\hat{k}$ are given. Can $\vec{c}$ be expressed as a linear combination of $\vec{a}$ and $\vec{b}$?

Scenario: Check if $\vec{c} = x\vec{a} + y\vec{b}$ for some scalars $x, y$. If a consistent solution for $x, y$ exists, then yes. If not, no.

$2\hat{i} + 0\hat{j} + 3\hat{k} = x(\hat{i} + \hat{j} + \hat{k}) + y(\hat{i} - \hat{j} + \hat{k}) = (x+y)\hat{i} + (x-y)\hat{j} + (x+y)\hat{k}$.

Equating coefficients: $x+y=2$ (from $\hat{i}$), $x-y=0$ (from $\hat{j}$), $x+y=3$ (from $\hat{k}$).

From $x-y=0$, $x=y$. Substituting into $x+y=2$ gives $2x=2 \Rightarrow x=1$, so $y=1$. But $x=1, y=1$ in $x+y=3$ gives $1+1=2 \neq 3$. Contradiction.

(A) Yes, $\vec{c} = \vec{a} + \vec{b}$.

(B) Yes, $\vec{c} = 1.5\vec{a} + 0.5\vec{b}$.

(C) No, because the system of equations for the scalar coefficients is inconsistent.

(D) Yes, because $\vec{a}$ and $\vec{b}$ are non-collinear.

Question 7. Find the unit vector in the direction of $\vec{v} = -6\hat{i} + 8\hat{j}$.

Scenario: A unit vector in the direction of $\vec{v}$ is $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$. First calculate the magnitude of $\vec{v}$.

$|\vec{v}| = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.

$\hat{v} = \frac{-6\hat{i} + 8\hat{j}}{10} = -\frac{6}{10}\hat{i} + \frac{8}{10}\hat{j} = -\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}$.

(A) $-\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}$

(B) $-6\hat{i} + 8\hat{j}$

(C) $\frac{1}{10}(-6\hat{i} + 8\hat{j})$

(D) $-\frac{6}{10}\hat{i} + \frac{8}{10}\hat{j}$

Question 1. A force $\vec{F} = 5\hat{i} + 2\hat{j}$ N is applied to an object, causing a displacement $\vec{d} = 3\hat{i} + 4\hat{j}$ m. Calculate the work done by the force.

Scenario: Work done is the scalar product of force and displacement: $W = \vec{F} \cdot \vec{d}$.

$W = (5\hat{i} + 2\hat{j}) \cdot (3\hat{i} + 4\hat{j}) = (5)(3) + (2)(4) = 15 + 8 = 23$ Joules.

(A) 23 J

(B) $15\hat{i} + 8\hat{j}$ J

(C) $\sqrt{29} \times 5$ J

(D) A vector quantity.

Question 2. Find the cosine of the angle between the vectors $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{j} + \hat{k}$.

Scenario: Use the formula $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$.

$\vec{a} \cdot \vec{b} = (1)(0) + (1)(1) + (0)(1) = 1$.

$|\vec{a}| = \sqrt{1^2+1^2+0^2} = \sqrt{2}$.

$|\vec{b}| = \sqrt{0^2+1^2+1^2} = \sqrt{2}$.

$\cos\theta = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2}$.

(A) 1

(B) 0

(C) $\frac{1}{2}$

(D) $\frac{1}{\sqrt{2}}$

Question 3. Check if the vectors $\vec{u} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{v} = \hat{i} + 5\hat{j} + \hat{k}$ are perpendicular.

Scenario: Two vectors are perpendicular if their dot product is zero.

$\vec{u} \cdot \vec{v} = (2)(1) + (-1)(5) + (3)(1) = 2 - 5 + 3 = 0$.

(A) Yes, because their dot product is 0.

(B) No, because their dot product is non-zero.

(C) Yes, because $\vec{u} \times \vec{v} \neq \vec{0}$.

(D) No, their cross product is zero.

Question 4. Find the scalar projection of the vector $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ on the vector $\vec{b} = 3\hat{i} - 4\hat{j}$.

Scenario: The scalar projection of $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.

$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-4) + (-2)(0) = 3 - 8 + 0 = -5$.

$|\vec{b}| = \sqrt{3^2 + (-4)^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

Scalar projection = $\frac{-5}{5} = -1$.

(A) 1

(B) -1

(C) $\frac{1}{5}(-5\hat{i})$

(D) A vector quantity.

Question 5. A box is pulled along a horizontal surface by a rope with a force of 50 N, inclined at an angle of $30^\circ$ to the horizontal. If the box is displaced by 10 m horizontally, calculate the work done by the rope force.

Scenario: Work done is $\vec{F} \cdot \vec{d} = |\vec{F}||\vec{d}|\cos\theta$. The displacement is horizontal, and the horizontal component of force does the work in this direction.

$W = (50 \text{ N})(10 \text{ m})\cos 30^\circ = 500 \times \frac{\sqrt{3}}{2} = 250\sqrt{3}$ J.

(A) 500 J

(B) 250 J

(C) $250\sqrt{3}$ J

(D) 0 J

Question 6. For non-zero vectors $\vec{a}$ and $\vec{b}$, if $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|$, what can be concluded about the angle between them?

Scenario: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$. If $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|$, then $\cos\theta = 1$.

(A) The angle is $90^\circ$.

(B) The angle is $0^\circ$.

(C) The vectors are perpendicular.

(D) The vectors are parallel and point in the same direction.

Question 1. Find the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a} = 2\hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - 3\hat{j}$.

Scenario: The area of the parallelogram is $|\vec{a} \times \vec{b}|$. First calculate the cross product.

$\vec{a} \times \vec{b} = (2\hat{i} + \hat{j}) \times (\hat{i} - 3\hat{j}) = 2(\hat{i} \times \hat{i}) - 6(\hat{i} \times \hat{j}) + (\hat{j} \times \hat{i}) - 3(\hat{j} \times \hat{j})$.

$= 2(\vec{0}) - 6(\hat{k}) + (-\hat{k}) - 3(\vec{0}) = -7\hat{k}$.

Area $= |-7\hat{k}| = 7$.

(A) 7 square units

(B) $-7\hat{k}$ square units

(C) $\sqrt{5} \times \sqrt{10}$ square units

(D) A vector quantity.

Question 2. A force $\vec{F} = 2\hat{j}$ N is applied at the point with position vector $\vec{r} = 3\hat{i}$ m relative to the origin. Calculate the torque about the origin.

Scenario: Torque is given by $\vec{\tau} = \vec{r} \times \vec{F}$.

$\vec{\tau} = (3\hat{i}) \times (2\hat{j}) = 6 (\hat{i} \times \hat{j}) = 6\hat{k}$ Nm.

(A) 6 Nm

(B) $6\hat{k}$ Nm

(C) 0 Nm

(D) $-6\hat{k}$ Nm

Question 3. Find a vector perpendicular to both $\vec{a} = \hat{i} - \hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$.

Scenario: The cross product of two vectors is perpendicular to both vectors.

$\vec{a} \times \vec{b} = (\hat{i} - \hat{j}) \times (\hat{i} + \hat{j}) = (\hat{i} \times \hat{i}) + (\hat{i} \times \hat{j}) - (\hat{j} \times \hat{i}) - (\hat{j} \times \hat{j})$.

$= \vec{0} + \hat{k} - (-\hat{k}) - \vec{0} = 2\hat{k}$. Any non-zero scalar multiple of $2\hat{k}$ is also perpendicular.

(A) $\hat{k}$

(B) $\hat{i}$

(C) $2\hat{k}$

(D) $\vec{0}$

Question 4. The angle between two non-zero vectors $\vec{a}$ and $\vec{b}$ is $\theta$. If $|\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b}$, what is the angle $\theta$?

Scenario: Substitute the definitions of magnitude of cross product and dot product and solve for $\theta$.

$|\vec{a}||\vec{b}|\sin\theta = |\vec{a}||\vec{b}|\cos\theta$. Assuming $|\vec{a}|, |\vec{b}| \neq 0$, $\sin\theta = \cos\theta$. For $\theta \in [0, \pi]$, this is true for $\theta = \pi/4$ or $45^\circ$.

(A) $0^\circ$

(B) $30^\circ$

(C) $45^\circ$

(D) $90^\circ$

Question 5. Find the area of the triangle with vertices P(1,2,3), Q(2,-1,1), R(0,1,2).

Scenario: The area of the triangle is $\frac{1}{2}|\vec{PQ} \times \vec{PR}|$. Calculate $\vec{PQ}$, $\vec{PR}$, their cross product, and the magnitude.

$\vec{PQ} = (2-1)\hat{i} + (-1-2)\hat{j} + (1-3)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k}$.

$\vec{PR} = (0-1)\hat{i} + (1-2)\hat{j} + (2-3)\hat{k} = -\hat{i} - \hat{j} - \hat{k}$.

$\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & -2 \\ -1 & -1 & -1 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(-1-2) + \hat{k}(-1-3) = \hat{i} + 3\hat{j} - 4\hat{k}$.

$|\vec{PQ} \times \vec{PR}| = \sqrt{1^2 + 3^2 + (-4)^2} = \sqrt{1 + 9 + 16} = \sqrt{26}$.

Area = $\frac{1}{2}\sqrt{26}$.

(A) $\frac{1}{2}\sqrt{26}$

(B) $\sqrt{26}$

(C) $\frac{1}{2}\sqrt{14}$

(D) 13

Question 6. For non-zero vectors $\vec{u}$ and $\vec{v}$, if $|\vec{u} \times \vec{v}| = 0$, what is the angle between them?

Scenario: $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin\theta$. If this is 0 and $|\vec{u}|, |\vec{v}| \neq 0$, then $\sin\theta = 0$.

(A) $90^\circ$

(B) $45^\circ$

(C) $0^\circ$ or $180^\circ$

(D) $60^\circ$

Question 7. A rod is pivoted at the origin. A force $\vec{F} = 5\hat{k}$ N is applied at the point with position vector $\vec{r} = 2\hat{i}$ m. Calculate the torque vector about the origin.

Scenario: Torque is $\vec{\tau} = \vec{r} \times \vec{F}$.

$\vec{\tau} = (2\hat{i}) \times (5\hat{k}) = 10 (\hat{i} \times \hat{k}) = 10(-\hat{j}) = -10\hat{j}$ Nm.

(A) $10\hat{j}$ Nm

(B) $-10\hat{j}$ Nm

(C) 10 Nm

(D) $\vec{0}$ Nm

Question 1. Find the volume of the parallelepiped whose adjacent edges are given by the vectors $\vec{a} = \hat{i}, \vec{b} = \hat{j}, \vec{c} = \hat{k}$.

Scenario: The volume of the parallelepiped is $|[\vec{a}, \vec{b}, \vec{c}]| = |\vec{a} \cdot (\vec{b} \times \vec{c})|$.

$\vec{b} \times \vec{c} = \hat{j} \times \hat{k} = \hat{i}$.

$\vec{a} \cdot (\vec{b} \times \vec{c}) = \hat{i} \cdot \hat{i} = 1$.

Volume = $|1| = 1$ cubic unit.

(A) 0

(B) 1

(C) $\sqrt{3}$

(D) 3

Question 2. Determine if the vectors $\vec{u} = \hat{i} - 2\hat{j} + \hat{k}$, $\vec{v} = 2\hat{i} + \hat{j} - 3\hat{k}$, and $\vec{w} = 3\hat{i} - 4\hat{j} + 5\hat{k}$ are coplanar.

Scenario: Three vectors are coplanar if their scalar triple product is zero.

$[\vec{u}, \vec{v}, \vec{w}] = \begin{vmatrix} 1 & -2 & 1 \\ 2 & 1 & -3 \\ 3 & -4 & 5 \end{vmatrix}$

$= 1(5 - 12) - (-2)(10 - (-9)) + 1(-8 - 3)$

$= 1(-7) + 2(19) + 1(-11) = -7 + 38 - 11 = 20 \neq 0$.

(A) Yes, because their scalar triple product is 0.

(B) No, because their scalar triple product is non-zero.

(C) Yes, because one vector is a linear combination of the others.

(D) Cannot be determined from the given information.

Question 3. The volume of the tetrahedron with vertices A(1,0,0), B(0,1,0), C(0,0,1), and D(1,1,1) needs to be calculated. What is the value?

Scenario: The volume of a tetrahedron with vertices P, Q, R, S is $\frac{1}{6} |[\vec{PQ}, \vec{PR}, \vec{PS}]|$. Let P=A.

$\vec{AB} = (0-1)\hat{i} + (1-0)\hat{j} + (0-0)\hat{k} = -\hat{i} + \hat{j}$.

$\vec{AC} = (0-1)\hat{i} + (0-0)\hat{j} + (1-0)\hat{k} = -\hat{i} + \hat{k}$.

$\vec{AD} = (1-1)\hat{i} + (1-0)\hat{j} + (1-0)\hat{k} = \hat{j} + \hat{k}$.

$[\vec{AB}, \vec{AC}, \vec{AD}] = \begin{vmatrix} -1 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix} = -1(0-1) - 1(-1-0) + 0(-1-0) = 1 + 1 + 0 = 2$.

Volume = $\frac{1}{6}|2| = \frac{2}{6} = \frac{1}{3}$.

(A) 2/3

(B) 1/6

(C) 1/3

(D) 2

Question 4. If the scalar triple product of three vectors $\vec{a}, \vec{b}, \vec{c}$ is 5, what is the volume of the parallelepiped formed by $2\vec{a}, \vec{b}, -\vec{c}$ as adjacent edges?

Scenario: $[k_1\vec{a}, k_2\vec{b}, k_3\vec{c}] = k_1k_2k_3[\vec{a}, \vec{b}, \vec{c}]$. Volume is the absolute value.

Volume = $|[2\vec{a}, \vec{b}, -\vec{c}]| = |2 \times 1 \times (-1) \times [\vec{a}, \vec{b}, \vec{c}]| = |-2 \times 5| = |-10| = 10$.

(A) 5

(B) 10

(C) -10

(D) 20

Question 5. For what value of $\lambda$ are the vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$, and $\vec{c} = \hat{i} + \lambda\hat{j} + 3\hat{k}$ coplanar?

Scenario: The vectors are coplanar if their scalar triple product is zero.

$[\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 1 & \lambda & 3 \end{vmatrix} = 0$.

$1(12 - \lambda) - 1(6 - 1) + 1(2\lambda - 4) = 0$.

$12 - \lambda - 5 + 2\lambda - 4 = 0$.

$\lambda + 3 = 0 \Rightarrow \lambda = -3$.

(A) 3

(B) -3

(C) 0

(D) 1

Question 6. If the scalar triple product $[\vec{a}, \vec{b}, \vec{c}]$ is negative, what does this imply about the orientation of the vectors?

Scenario: The sign of the scalar triple product indicates whether the vectors form a right-handed or left-handed system.

(A) They are coplanar.

(B) They form a right-handed system.

(C) They form a left-handed system.

(D) The volume of the parallelepiped is negative (which is not possible for volume).

Question 1. Point A has position vector $2\hat{i} + \hat{j} - \hat{k}$ and point B has position vector $\hat{i} - 2\hat{j} + 3\hat{k}$. Find the position vector of point R that divides AB internally in the ratio 1:2.

Scenario: Apply the internal section formula with $m=1, n=2, \vec{a} = 2\hat{i} + \hat{j} - \hat{k}, \vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$.

$\vec{r} = \frac{2\vec{a} + 1\vec{b}}{1+2} = \frac{2(2\hat{i} + \hat{j} - \hat{k}) + (\hat{i} - 2\hat{j} + 3\hat{k})}{3}$

$= \frac{4\hat{i} + 2\hat{j} - 2\hat{k} + \hat{i} - 2\hat{j} + 3\hat{k}}{3} = \frac{5\hat{i} + \hat{k}}{3} = \frac{5}{3}\hat{i} + \frac{1}{3}\hat{k}$.

(A) $\frac{1}{3}(5\hat{i} + \hat{k})$

(B) $5\hat{i} + \hat{k}$

(C) $\frac{1}{3}(4\hat{i} + \hat{j} + 5\hat{k})$

(D) $\frac{1}{3}(\hat{i} - 5\hat{j} + 7\hat{k})$

Question 2. Find the position vector of the midpoint of the line segment joining the points P(1, -1, 2) and Q(3, 5, -4).

Scenario: The midpoint is the point that divides the segment internally in the ratio 1:1. Apply the midpoint formula or section formula with m=n=1.

Position vector of midpoint $= \frac{(1\hat{i} - 1\hat{j} + 2\hat{k}) + (3\hat{i} + 5\hat{j} - 4\hat{k})}{2}$

$= \frac{(1+3)\hat{i} + (-1+5)\hat{j} + (2-4)\hat{k}}{2} = \frac{4\hat{i} + 4\hat{j} - 2\hat{k}}{2} = 2\hat{i} + 2\hat{j} - \hat{k}$.

(A) $(2, 2, -1)$

(B) $4\hat{i} + 4\hat{j} - 2\hat{k}$

(C) $2\hat{i} + 2\hat{j} - \hat{k}$

(D) $\hat{i} + 3\hat{j} - 3\hat{k}$

Question 3. The position vectors of points A and B are $\vec{a}$ and $\vec{b}$. A point R divides AB externally in the ratio 2:1. Find the position vector of R.

Scenario: Apply the external section formula with $m=2, n=1, \vec{a}, \vec{b}$.

$\vec{r} = \frac{m\vec{b} - n\vec{a}}{m-n} = \frac{2\vec{b} - 1\vec{a}}{2-1} = 2\vec{b} - \vec{a}$.

(A) $\frac{2\vec{b} + \vec{a}}{3}$

(B) $2\vec{b} - \vec{a}$

(C) $\vec{a} - 2\vec{b}$

(D) $\frac{\vec{b} - 2\vec{a}}{-1}$

Question 4. The vertices of a triangle are A(1,1,1), B(2,3,4), C(3,-1,2). Find the position vector of its centroid.

Scenario: The position vector of the centroid of a triangle is the average of the position vectors of its vertices.

Position vector of centroid $= \frac{(1\hat{i}+\hat{j}+\hat{k}) + (2\hat{i}+3\hat{j}+4\hat{k}) + (3\hat{i}-\hat{j}+2\hat{k})}{3}$

$= \frac{(1+2+3)\hat{i} + (1+3-1)\hat{j} + (1+4+2)\hat{k}}{3} = \frac{6\hat{i} + 3\hat{j} + 7\hat{k}}{3} = 2\hat{i} + \hat{j} + \frac{7}{3}\hat{k}$.

(A) $6\hat{i} + 3\hat{j} + 7\hat{k}$

(B) $(2, 1, \frac{7}{3})$

(C) $\frac{1}{3}(6, 3, 7)$

(D) $\frac{1}{3}(\vec{a}+\vec{b}+\vec{c})$

Question 5. Point R divides the line segment PQ in the ratio 3:2 externally. If the position vector of P is $\vec{p}$ and R is $\vec{r}$, express the position vector of Q ($\vec{q}$) in terms of $\vec{p}$ and $\vec{r}$.

Scenario: R divides PQ externally in ratio $m:n$, so $\vec{r} = \frac{m\vec{q} - n\vec{p}}{m-n}$. Substitute $m=3, n=2$ and rearrange to find $\vec{q}$.

$\vec{r} = \frac{3\vec{q} - 2\vec{p}}{3-2} = 3\vec{q} - 2\vec{p}$.

$3\vec{q} = \vec{r} + 2\vec{p}$.

$\vec{q} = \frac{1}{3}(\vec{r} + 2\vec{p})$.

(A) $\frac{1}{3}(\vec{r} + 2\vec{p})$

(B) $3\vec{r} - 2\vec{p}$

(C) $\frac{1}{3}(3\vec{r} + 2\vec{p})$

(D) $\frac{1}{3}(\vec{r} - 2\vec{p})$

Question 6. Let O be the origin. If point P divides the line segment AB internally in the ratio $m:n$, where A and B have position vectors $\vec{a}$ and $\vec{b}$ respectively, then the vector $\vec{AP}$ is:

Scenario: $\vec{AP} = \vec{OP} - \vec{OA}$. Use the internal section formula for $\vec{OP}$ (or $\vec{r}$).

$\vec{OP} = \frac{n\vec{a} + m\vec{b}}{m+n}$.

$\vec{AP} = \frac{n\vec{a} + m\vec{b}}{m+n} - \vec{a} = \frac{n\vec{a} + m\vec{b} - (m+n)\vec{a}}{m+n} = \frac{n\vec{a} + m\vec{b} - m\vec{a} - n\vec{a}}{m+n} = \frac{m\vec{b} - m\vec{a}}{m+n} = \frac{m(\vec{b}-\vec{a})}{m+n}$.

(A) $\frac{n(\vec{b}-\vec{a})}{m+n}$

(B) $\frac{m(\vec{b}-\vec{a})}{m+n}$

(C) $\frac{m\vec{b} + n\vec{a}}{m+n}$

(D) $\frac{m}{m+n}\vec{AB}$

Question 1. Find the distance of the point (3, -2, 5) from the xz-plane.

Scenario: The xz-plane is defined by $y=0$. The distance of a point $(x_1, y_1, z_1)$ from a plane is the length of the perpendicular from the point to the plane. For a plane $y=0$, the distance is $|y_1|$.

(A) 3

(B) -2

(C) 2

(D) 5

Question 2. A line passes through the point (1, 2, -1) and has direction ratios (2, 3, -4). Write the vector equation of this line.

Scenario: The vector equation of a line passing through $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r} = \vec{a} + \lambda \vec{b}$.

$\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$. The direction ratios $(2, 3, -4)$ correspond to the direction vector $\vec{b} = 2\hat{i} + 3\hat{j} - 4\hat{k}$.

(A) $\vec{r} = (2\hat{i} + 3\hat{j} - 4\hat{k}) + \lambda (\hat{i} + 2\hat{j} - \hat{k})$

(B) $\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda (2\hat{i} + 3\hat{j} - 4\hat{k})$

(C) $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+1}{-4}$

(D) $\vec{r} = (1+\lambda 2)\hat{i} + (2+\lambda 3)\hat{j} + (-1+\lambda (-4))\hat{k}$

Question 3. Find the Cartesian equation of the line passing through points A(1, 2, -1) and B(3, -1, 2).

Scenario: The Cartesian equation of a line passing through $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.

$(x_1, y_1, z_1) = (1, 2, -1)$, $(x_2, y_2, z_2) = (3, -1, 2)$.

$x_2-x_1 = 2$, $y_2-y_1 = -3$, $z_2-z_1 = 3$.

(A) $\frac{x-1}{2} = \frac{y-2}{-3} = \frac{z+1}{3}$

(B) $\frac{x-3}{1} = \frac{y+1}{2} = \frac{z-2}{-1}$

(C) $\frac{x-1}{3} = \frac{y-2}{-1} = \frac{z+1}{2}$

(D) $\frac{x+1}{2} = \frac{y+2}{-3} = \frac{z-1}{3}$

Question 4. A line in 3D space makes angles $60^\circ, 45^\circ, 60^\circ$ with the positive x, y, and z axes respectively. What are its direction cosines?

Scenario: The direction cosines are the cosines of the angles the line makes with the positive coordinate axes.

$\ell = \cos 60^\circ = 1/2$.

$m = \cos 45^\circ = 1/\sqrt{2}$.

$n = \cos 60^\circ = 1/2$.

Check if $\ell^2 + m^2 + n^2 = 1$: $(1/2)^2 + (1/\sqrt{2})^2 + (1/2)^2 = 1/4 + 1/2 + 1/4 = 1/4 + 2/4 + 1/4 = 4/4 = 1$. So these are valid direction cosines.

(A) $(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2})$

(B) $(\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}, \frac{\sqrt{3}}{2})$

(C) $(1/2, 1/\sqrt{2}, 1/2)$

(D) $(0.5, 0.707, 0.5)$ (approx values)

Question 5. The Cartesian equation of a line is $\frac{x-5}{3} = \frac{y+4}{2} = \frac{z-1}{-1}$. Find the vector equation of this line.

Scenario: From the Cartesian equation, identify a point on the line $(x_1, y_1, z_1)$ and the direction ratios $(a, b, c)$. Convert these to vector form.

Point is $(5, -4, 1)$, so $\vec{a} = 5\hat{i} - 4\hat{j} + \hat{k}$.

Direction ratios are $(3, 2, -1)$, so direction vector $\vec{b} = 3\hat{i} + 2\hat{j} - \hat{k}$.

(A) $\vec{r} = (3\hat{i} + 2\hat{j} - \hat{k}) + \lambda (5\hat{i} - 4\hat{j} + \hat{k})$

(B) $\vec{r} = (5\hat{i} + 4\hat{j} - \hat{k}) + \lambda (3\hat{i} + 2\hat{j} - \hat{k})$

(C) $\vec{r} = (5\hat{i} - 4\hat{j} + \hat{k}) + \lambda (3\hat{i} + 2\hat{j} - \hat{k})$

(D) $\vec{r} = (5\hat{i} - 4\hat{j} + \hat{k}) + \lambda (-3\hat{i} - 2\hat{j} + \hat{k})$

Question 6. Find the coordinates of a point on the line $\vec{r} = (1\hat{i} - 3\hat{j} + 2\hat{k}) + \lambda (-2\hat{i} + \hat{j} + \hat{k})$ when $\lambda = 2$.

Scenario: Substitute the value of the parameter $\lambda$ into the vector equation and read the components.

$\vec{r} = (\hat{i} - 3\hat{j} + 2\hat{k}) + 2 (-2\hat{i} + \hat{j} + \hat{k})$

$= \hat{i} - 3\hat{j} + 2\hat{k} - 4\hat{i} + 2\hat{j} + 2\hat{k}$

$= (1-4)\hat{i} + (-3+2)\hat{j} + (2+2)\hat{k} = -3\hat{i} - \hat{j} + 4\hat{k}$.

Coordinates are $(-3, -1, 4)$.

(A) $(-3, -1, 4)$

(B) $(1, -3, 2)$

(C) $(-2, 1, 1)$

(D) $(-4, 2, 2)$

Question 1. Write the vector equation (normal form) of the plane $2x - 3y + 6z = 14$.

Scenario: Convert the Cartesian equation $Ax+By+Cz+D=0$ (or $Ax+By+Cz=k$) to vector form $\vec{r} \cdot \hat{n} = d$. Find the normal vector, its magnitude, the unit normal, and the distance from origin.

Normal vector $\vec{n} = 2\hat{i} - 3\hat{j} + 6\hat{k}$.

$|\vec{n}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

Unit normal $\hat{n} = \frac{1}{7}(2\hat{i} - 3\hat{j} + 6\hat{k})$.

Distance from origin $d = \frac{|14|}{|\vec{n}|} = \frac{14}{7} = 2$.

(A) $\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 14$

(B) $\vec{r} \cdot \frac{1}{7}(2\hat{i} - 3\hat{j} + 6\hat{k}) = 14$

(C) $\vec{r} \cdot \frac{1}{7}(2\hat{i} - 3\hat{j} + 6\hat{k}) = 2$

(D) $\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 2$

Question 2. Find the equation of the plane passing through the point (1, 2, -1) and perpendicular to the vector $3\hat{i} + \hat{j} - 2\hat{k}$.

Scenario: Use the point-normal form of the plane equation, $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$, where $\vec{a}$ is the position vector of the point and $\vec{n}$ is the normal vector. Convert to Cartesian form.

$\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$, $\vec{n} = 3\hat{i} + \hat{j} - 2\hat{k}$.

$(\vec{r} - (\hat{i} + 2\hat{j} - \hat{k})) \cdot (3\hat{i} + \hat{j} - 2\hat{k}) = 0$.

$(x\hat{i} + y\hat{j} + z\hat{k} - \hat{i} - 2\hat{j} + \hat{k}) \cdot (3\hat{i} + \hat{j} - 2\hat{k}) = 0$.

$((x-1)\hat{i} + (y-2)\hat{j} + (z+1)\hat{k}) \cdot (3\hat{i} + \hat{j} - 2\hat{k}) = 0$.

$3(x-1) + 1(y-2) - 2(z+1) = 0$.

$3x - 3 + y - 2 - 2z - 2 = 0$.

$3x + y - 2z - 7 = 0$.

(A) $3x + y - 2z = 0$

(B) $3x + y - 2z = 7$

(C) $x + 2y - z = 7$

(D) $3x + y - 2z - 7 = 0$

Question 3. Find the equation of the plane that makes intercepts 2, 3, and 4 on the x, y, and z axes respectively.

Scenario: Use the intercept form of the plane equation $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a, b, c$ are the intercepts.

$a=2, b=3, c=4$.

$\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$. Multiply by the LCM of 2, 3, 4, which is 12.

$6x + 4y + 3z = 12$.

(A) $2x + 3y + 4z = 1$

(B) $6x + 4y + 3z = 12$

(C) $12x + 8y + 6z = 1$

(D) $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 0$

Question 4. A plane passes through the point (1, -1, 2) and is parallel to the plane $x - 2y + 3z = 1$. Find the equation of this plane.

Scenario: Parallel planes have the same normal vector. The normal vector of the given plane is $\hat{i} - 2\hat{j} + 3\hat{k}$. Use the point-normal form with the given point and this normal vector.

Normal vector $\vec{n} = \hat{i} - 2\hat{j} + 3\hat{k}$. Point $(x_1, y_1, z_1) = (1, -1, 2)$.

$1(x-1) - 2(y-(-1)) + 3(z-2) = 0$.

$x - 1 - 2(y+1) + 3z - 6 = 0$.

$x - 1 - 2y - 2 + 3z - 6 = 0$.

$x - 2y + 3z - 9 = 0$.

(A) $x - 2y + 3z = 1$

(B) $x - 2y + 3z = -9$

(C) $x - 2y + 3z = 9$

(D) $x - 2y + 3z - 9 = 0$

Question 5. Find the equation of the plane passing through the intersection of the planes $x+y+z = 6$ and $2x+3y+4z = 5$, and passing through the point (1,1,1).

Scenario: The equation of a plane through the intersection of $P_1=0$ and $P_2=0$ is $P_1 + \lambda P_2 = 0$. Substitute the point (1,1,1) to find $\lambda$.

$(x+y+z-6) + \lambda(2x+3y+4z-5) = 0$.

Substitute (1,1,1): $(1+1+1-6) + \lambda(2(1)+3(1)+4(1)-5) = 0$.

$(-3) + \lambda(2+3+4-5) = 0$.

$-3 + \lambda(4) = 0 \Rightarrow 4\lambda = 3 \Rightarrow \lambda = 3/4$.

Substitute $\lambda = 3/4$ back into the plane equation.

$(x+y+z-6) + \frac{3}{4}(2x+3y+4z-5) = 0$. Multiply by 4.

$4(x+y+z-6) + 3(2x+3y+4z-5) = 0$.

$4x+4y+4z-24 + 6x+9y+12z-15 = 0$.

$10x + 13y + 16z - 39 = 0$.

(A) $10x + 13y + 16z = 39$

(B) $10x + 13y + 16z + 39 = 0$

(C) $10x + 13y + 16z = 0$

(D) $4x+4y+4z-24 + 6x+9y+12z-15 = 0$

Question 6. What is the normal vector to the plane $\vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) = 5$?

Scenario: In the vector form $\vec{r} \cdot \vec{n} = d$, $\vec{n}$ is the normal vector.

(A) $2\hat{i} + \hat{j} - \hat{k}$

(B) $5$

(C) $\frac{1}{\sqrt{6}}(2\hat{i} + \hat{j} - \hat{k})$

(D) A vector parallel to the plane.

Question 7. Find the equation of the plane passing through the points P(1, -1, 1), Q(1, 1, -1), and R(-1, 1, 1).

Scenario: Find two vectors in the plane, e.g., $\vec{PQ}$ and $\vec{PR}$. Their cross product is normal to the plane. Use one point and the normal vector in the point-normal form.

$\vec{PQ} = (1-1)\hat{i} + (1-(-1))\hat{j} + (-1-1)\hat{k} = 2\hat{j} - 2\hat{k}$.

$\vec{PR} = (-1-1)\hat{i} + (1-(-1))\hat{j} + (1-1)\hat{k} = -2\hat{i} + 2\hat{j}$.

$\vec{n} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -2 \\ -2 & 2 & 0 \end{vmatrix} = \hat{i}(0 - (-4)) - \hat{j}(0 - 4) + \hat{k}(0 - (-4)) = 4\hat{i} + 4\hat{j} + 4\hat{k}$.

We can use $\vec{n} = \hat{i} + \hat{j} + \hat{k}$ as the normal vector (proportional). Using point P(1,-1,1):

$1(x-1) + 1(y-(-1)) + 1(z-1) = 0$.

$x - 1 + y + 1 + z - 1 = 0$.

$x + y + z - 1 = 0$.

(A) $x + y + z = 1$

(B) $x + y + z = 3$

(C) $4x + 4y + 4z = 4$

(D) $x + y + z - 1 = 0$

Question 1. Find the cosine of the angle between the lines $\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda (2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \mu (\hat{i} + \hat{j} + 2\hat{k})$.

Scenario: The angle between the lines is the angle between their direction vectors $\vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b}_2 = \hat{i} + \hat{j} + 2\hat{k}$. Use $\cos\theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1||\vec{b}_2|}$.

$\vec{b}_1 \cdot \vec{b}_2 = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$.

$|\vec{b}_1| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$.

$|\vec{b}_2| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1+1+4} = \sqrt{6}$.

$\cos\theta = \frac{|3|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.

(A) $\frac{1}{2}$

(B) $-\frac{1}{2}$

(C) 0

(D) 1

Question 2. Find the angle between the planes $x+y+z=1$ and $2x+y-z=3$.

Scenario: The angle between the planes is the angle between their normal vectors $\vec{n}_1 = \hat{i} + \hat{j} + \hat{k}$ and $\vec{n}_2 = 2\hat{i} + \hat{j} - \hat{k}$. Use $\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|}$.

$\vec{n}_1 \cdot \vec{n}_2 = (1)(2) + (1)(1) + (1)(-1) = 2 + 1 - 1 = 2$.

$|\vec{n}_1| = \sqrt{1^2+1^2+1^2} = \sqrt{3}$.

$|\vec{n}_2| = \sqrt{2^2+1^2+(-1)^2} = \sqrt{4+1+1} = \sqrt{6}$.

$\cos\theta = \frac{|2|}{\sqrt{3}\sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}$.

(A) $\cos^{-1}(\frac{2}{\sqrt{18}})$

(B) $\cos^{-1}(\frac{\sqrt{2}}{3})$

(C) $\cos^{-1}(\frac{2}{3})$

(D) $\cos^{-1}(2)$

Question 3. Determine if the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ is parallel to the plane $4x + 6y - 10z + 7 = 0$.

Scenario: A line with direction vector $\vec{b}$ is parallel to a plane with normal vector $\vec{n}$ if $\vec{b} \cdot \vec{n} = 0$.

Direction vector of line $\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

Normal vector of plane $\vec{n} = 4\hat{i} + 6\hat{j} - 10\hat{k}$.

$\vec{b} \cdot \vec{n} = (2)(4) + (3)(6) + (4)(-10) = 8 + 18 - 40 = 26 - 40 = -14$.

Since $\vec{b} \cdot \vec{n} = -14 \neq 0$, the line is NOT parallel to the plane. The angle between the line and the normal is not 90 degrees.

(A) Yes, because the dot product of direction ratios and normal coefficients is 0.

(B) No, because the dot product of the direction vector and normal vector is non-zero.

(C) Yes, because the cross product of the direction vector and normal vector is zero.

(D) No, because the line intersects the plane.

Question 4. Find the sine of the angle between the line $\vec{r} = (2\hat{i} + \hat{j} - 3\hat{k}) + \lambda (\hat{i} + 2\hat{j} - 2\hat{k})$ and the plane $\vec{r} \cdot (2\hat{i} - \hat{j} + 4\hat{k}) = 5$.

Scenario: The angle $\phi$ between the line (direction $\vec{b}$) and the plane (normal $\vec{n}$) is given by $\sin\phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}$.

Direction vector of line $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$.

Normal vector of plane $\vec{n} = 2\hat{i} - \hat{j} + 4\hat{k}$.

$\vec{b} \cdot \vec{n} = (1)(2) + (2)(-1) + (-2)(4) = 2 - 2 - 8 = -8$.

$|\vec{b}| = \sqrt{1^2+2^2+(-2)^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.

$|\vec{n}| = \sqrt{2^2+(-1)^2+4^2} = \sqrt{4+1+16} = \sqrt{21}$.

$\sin\phi = \frac{|-8|}{3\sqrt{21}} = \frac{8}{3\sqrt{21}}$.

(A) $\frac{8}{3\sqrt{21}}$

(B) $\frac{|-8|}{\sqrt{9}\sqrt{21}}$

(C) $\frac{-8}{3\sqrt{21}}$

(D) $\frac{8}{\sqrt{21}}$

Question 5. Check if the planes $3x - 2y + z = 4$ and $6x - 4y + 2z = 5$ are parallel.

Scenario: Two planes are parallel if their normal vectors are parallel (components proportional).

Normal vector 1: $\vec{n}_1 = 3\hat{i} - 2\hat{j} + \hat{k}$.

Normal vector 2: $\vec{n}_2 = 6\hat{i} - 4\hat{j} + 2\hat{k}$.

Check proportionality: $\frac{6}{3} = 2$, $\frac{-4}{-2} = 2$, $\frac{2}{1} = 2$. The components are proportional. The normals are parallel. Also check the constant terms: $\frac{5}{4} \neq 2$. Since the ratio of constants is different from the ratio of coefficients, the planes are parallel and distinct.

(A) Yes, because their normal vectors are proportional, and the ratio of constant terms is different.

(B) No, because the ratio of constant terms is different.

(C) Yes, because their normal vectors are parallel.

(D) No, their normal vectors' dot product is non-zero.

Question 6. Find the angle between the line $\vec{r} = \lambda \hat{k}$ (z-axis) and the line $\vec{r} = \mu \hat{j}$ (y-axis).

Scenario: The angle between lines is the angle between their direction vectors. The direction vector of the first line is $\hat{k}$, and the direction vector of the second line is $\hat{j}$.

Angle is the angle between $\hat{k}$ and $\hat{j}$. These are orthogonal unit vectors.

(A) $0^\circ$

(B) $45^\circ$

(C) $90^\circ$

(D) $180^\circ$

Question 1. Find the distance of the point P(1, -2, 3) from the plane $x+2y-2z+5=0$.

Scenario: Use the point-to-plane distance formula $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2+B^2+C^2}}$.

$(x_1, y_1, z_1) = (1, -2, 3)$, $A=1, B=2, C=-2, D=5$.

$d = \frac{|1(1) + 2(-2) + (-2)(3) + 5|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|1 - 4 - 6 + 5|}{\sqrt{1 + 4 + 4}} = \frac{|-4|}{\sqrt{9}} = \frac{4}{3}$.

(A) $\frac{4}{3}$

(B) $\frac{4}{\sqrt{3}}$

(C) $\frac{1}{3}$

(D) 4

Question 2. Find the shortest distance between the parallel lines $\vec{r} = \hat{i} + \hat{j} + \lambda (2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu (2\hat{i} - \hat{j} + \hat{k})$.

Scenario: Use the formula for shortest distance between parallel lines $d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$.

$\vec{a}_1 = \hat{i} + \hat{j}$, $\vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$.

$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (1-1)\hat{j} + (-1-0)\hat{k} = \hat{i} - \hat{k}$.

$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = (\hat{i} - \hat{k}) \times (2\hat{i} - \hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -1 \\ 2 & -1 & 1 \end{vmatrix}$

$= \hat{i}(0 - 1) - \hat{j}(1 - (-2)) + \hat{k}(-1 - 0) = -\hat{i} - 3\hat{j} - \hat{k}$.

$|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{(-1)^2 + (-3)^2 + (-1)^2} = \sqrt{1 + 9 + 1} = \sqrt{11}$.

$|\vec{b}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.

$d = \frac{\sqrt{11}}{\sqrt{6}} = \sqrt{\frac{11}{6}}$.

(A) $\sqrt{\frac{11}{6}}$

(B) $\frac{\sqrt{11}}{6}$

(C) $\frac{11}{6}$

(D) $\sqrt{11}$

Question 3. Calculate the shortest distance between the skew lines $\vec{r} = \hat{i} + \hat{j} + \lambda (2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} - \hat{j} + \hat{k} + \mu (3\hat{i} - 5\hat{j} + 2\hat{k})$.

Scenario: Use the formula for shortest distance between skew lines $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.

$\vec{a}_1 = \hat{i} + \hat{j}$, $\vec{a}_2 = 2\hat{i} - \hat{j} + \hat{k}$, $\vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k}$, $\vec{b}_2 = 3\hat{i} - 5\hat{j} + 2\hat{k}$.

$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (-1-1)\hat{j} + (1-0)\hat{k} = \hat{i} - 2\hat{j} + \hat{k}$.

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} = \hat{i}(-2 - (-5)) - \hat{j}(4 - 3) + \hat{k}(-10 - (-3)) = \hat{i}(3) - \hat{j}(1) + \hat{k}(-7) = 3\hat{i} - \hat{j} - 7\hat{k}$.

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9 + 1 + 49} = \sqrt{59}$.

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (\hat{i} - 2\hat{j} + \hat{k}) \cdot (3\hat{i} - \hat{j} - 7\hat{k})$

$= (1)(3) + (-2)(-1) + (1)(-7) = 3 + 2 - 7 = -2$.

$d = \frac{|-2|}{\sqrt{59}} = \frac{2}{\sqrt{59}}$.

(A) $\frac{2}{\sqrt{59}}$

(B) $\frac{|-2|}{\sqrt{59}}$

(C) 0

(D) $\frac{2}{\sqrt{11}\sqrt{59}}$

Question 4. Find the distance between the parallel planes $x+2y-2z=3$ and $3x+6y-6z=15$.

Scenario: First, ensure the normal vectors are identical or proportional. The second equation is $3(x+2y-2z) = 15$, so $x+2y-2z = 5$. The planes are parallel with normal vector $\hat{i} + 2\hat{j} - 2\hat{k}$. Use the formula for distance between parallel planes $d = \frac{|D_2 - D_1|}{\sqrt{A^2+B^2+C^2}}$.

Plane 1: $x+2y-2z-3=0 \Rightarrow A=1, B=2, C=-2, D_1=-3$.

Plane 2: $x+2y-2z-5=0 \Rightarrow A=1, B=2, C=-2, D_2=-5$.

$\sqrt{A^2+B^2+C^2} = \sqrt{1^2+2^2+(-2)^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.

$d = \frac{|-5 - (-3)|}{3} = \frac{|-2|}{3} = \frac{2}{3}$.

(A) $\frac{|15-3|}{\sqrt{3^2+6^2+(-6)^2}}$

(B) $\frac{|15-9|}{\sqrt{3^2+6^2+(-6)^2}} = \frac{6}{\sqrt{81}} = \frac{6}{9} = \frac{2}{3}$

(C) $\frac{2}{3}$

(D) 0

Question 5. Find the shortest distance between the line passing through the origin with direction $(1,1,1)$ and the line passing through $(1,0,0)$ with direction $(1,1,1)$.

Scenario: Both lines have the same direction vector $(1,1,1)$, so they are parallel. Use the formula for distance between parallel lines.

Line 1: $\vec{r} = \vec{0} + \lambda (\hat{i} + \hat{j} + \hat{k})$. $\vec{a}_1 = \vec{0}$.

Line 2: $\vec{r} = \hat{i} + \mu (\hat{i} + \hat{j} + \hat{k})$. $\vec{a}_2 = \hat{i}$.

Common direction vector $\vec{b} = \hat{i} + \hat{j} + \hat{k}$.

$\vec{a}_2 - \vec{a}_1 = \hat{i} - \vec{0} = \hat{i}$.

$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \hat{i} \times (\hat{i} + \hat{j} + \hat{k}) = (\hat{i} \times \hat{i}) + (\hat{i} \times \hat{j}) + (\hat{i} \times \hat{k}) = \vec{0} + \hat{k} - \hat{j} = -\hat{j} + \hat{k}$.

$|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = |-\hat{j} + \hat{k}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.

$|\vec{b}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}$.

$d = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}$.

(A) 0

(B) 1

(C) $\sqrt{\frac{2}{3}}$

(D) $\frac{\sqrt{2}}{3}$