Target Exam

CUET

Subject

Section B1

Chapter

Vectors

Question:

Find a unit vector in the direction of $\vec{PQ}$, where $P$ and $Q$ have coordinates $(5, 0, 8)$ and $(3, 3, 2)$, respectively.

Options:

$\frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k}$

$-\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}$

$-2\hat{i} + 3\hat{j} - 6\hat{k}$

$-\frac{2}{49}\hat{i} + \frac{3}{49}\hat{j} - \frac{6}{49}\hat{k}$

Correct Answer:

$-\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}$

Explanation:

The correct answer is Option (2) → $-\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}$ ##

Since, the coordinates of $P$ and $Q$ are $(5, 0, 8)$ and $(3, 3, 2)$, respectively.

$∴\vec{PQ} = \vec{OQ} - \vec{OP}$

$= (3\hat{i} + 3\hat{j} + 2\hat{k}) - (5\hat{i} + 0\hat{j} + 8\hat{k})$

$= -2\hat{i} + 3\hat{j} - 6\hat{k}$

$∴$ Unit vector in the direction of $\vec{PQ} = \frac{\vec{PQ}}{|\vec{PQ}|}$

$= \frac{-2\hat{i} + 3\hat{j} - 6\hat{k}}{\sqrt{(-2)^2 + (3)^2 + (-6)^2}}$

$= \frac{-2\hat{i} + 3\hat{j} - 6\hat{k}}{\sqrt{49}} = \frac{-2\hat{i} + 3\hat{j} - 6\hat{k}}{7}$