If the vectors a = 3i - pj + 5k and b = -6i + 14j + qk are collinear, then the value of p and q are: Option 1: p = -7, q = 18 Option 2: p = 7, q = -20 Option 3: p = 7, q = -10 Option 4: p = -7, q = 5

Answer: p = 7, q = -10. Solution: Given: a = 3i - pj + 5k b = -6i + 14j + qk The vectors are collinear. For collinear vectors, b = ka where k is a constant. Using the i component: -6 = k x 3 k = (-6)/3 k = -2 Using the j component: 14 = k x (-p) 14 = (-2) x (-p) 14 = 2p p = 14/2 p = 7 Using the k component: q = k x 5 q = (-2) x 5 q = -10 Therefore, p = 7 and q = -10.

Algebra

Vector Algebra

Medium

core

If the vectors $\vec{a} = 3\hat{i} - p\hat{j} + 5\hat{k}$ and $\vec{b} = -6\hat{i} + 14\hat{j} + q\hat{k}$ are collinear, then the value of p and q are:

$p = -7, q = 18$

$p = 7, q = -20$

$p = 7, q = -10$

$p = -7, q = 5$

✅ Correct Option: 3

Given:

$\vec{a} = 3\hat{i} - p\hat{j} + 5\hat{k}$

$\vec{b} = -6\hat{i} + 14\hat{j} + q\hat{k}$

The vectors are collinear.

For collinear vectors, $\vec{b} = k\vec{a}$ where $k$ is a constant.

Using the $\hat{i}$ component:

$-6 = k \times 3$

$k = \frac{-6}{3}$

$k = -2$

Using the $\hat{j}$ component:

$14 = k \times (-p)$

$14 = (-2) \times (-p)$

$14 = 2p$

$p = \frac{14}{2}$

$p = 7$

Using the $\hat{k}$ component:

$q = k \times 5$

$q = (-2) \times 5$

$q = -10$

Therefore, $p = 7$ and $q = -10$ .

2023: 23 May Shift 3

The unit vector in the direction of $\vec{a} + \vec{b}$ if $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$ & $\vec{b} = -\hat{i} + \hat{j} + -\hat{k}$ is :

2026: 29 May Shift 1

If a unit vector $\hat a$ makes an angle $\dfrac{\pi}{3}$ with $\hat i$ , $\dfrac{\pi}{4}$ with $\hat j$ and an acute angle $\theta$ with $\hat k$ , then the respective value of $\theta$ and $\hat a$ are

2026: 15 May Shift 1

If $|\vec a|=2, |\vec b|=7, \vec a \times \vec b = 3\hat i + 2\hat j + 6\hat k$ , then the angle between $\vec a$ and $\vec b$ is equal to

2026: 31 May Shift 2

In $\triangle ABC$ , Which of the following are True?

A. $\overline{AB}+\overline{BC}+\overline{CA}=\vec{0}$

B. $\overline{AB}+\overline{BC}-\overline{AC}=\vec{0}$

C. $\overline{AB}+\overline{BC}-\overline{CA}=\vec{0}$

D. $\overline{AB}-\overline{CB}+\overline{CA}=\vec{0}$

Choose the correct answer from the options given below:

2022: 30 Aug Shift 1

If $\vec{p} = \hat{i} + \hat{j} - 2\hat{k}$ and $\vec{q} = 2\hat{i} + \hat{j} - \hat{k}$ , then the area of parallelogram having diagonals $(\vec{p} + \vec{q})$ and $(\vec{p} - \vec{q})$ is

2025: 2 June Shift 1

Let $\vec{a}$ and $\vec{b}$ are unit vectors. If $\sqrt{3}\vec{a} - \vec{b}$ is a unit vector, then the angle between $\vec{a}$ and $\vec{b}$ is

If the vectors a⃗=3i^−pj^+5k^\vec{a} = 3\hat{i} - p\hat{j} + 5\hat{k}a=3i^−pj^​+5k^ and b⃗=−6i^+14j^+qk^\vec{b} = -6\hat{i} + 14\hat{j} + q\hat{k}b=−6i^+14j^​+qk^ are collinear, then the value of p and q are:

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

If the vectors $\vec{a} = 3\hat{i} - p\hat{j} + 5\hat{k}$ and $\vec{b} = -6\hat{i} + 14\hat{j} + q\hat{k}$ are collinear, then the value of p and q are: