CBSE Sample Paper 2020-21 · Part A: Section I · 1 mark

Q10: Unit vector opposite to $-\frac{3}{4}\hat j$

The question

Find a unit vector in the direction opposite to $-\frac{3}{4}\hat j$

Exam answer

The vector opposite to $-\frac{3}{4}\hat j$ is $\frac{3}{4}\hat j$ ⁠.

Why?

Multiplying a vector by $-1$ reverses its direction: $-\left(-\frac{3}{4}\hat j\right) = \frac{3}{4}\hat j$ ⁠.

\text{Unit vector} = \frac{\frac{3}{4}\hat j}{\left|\frac{3}{4}\hat j\right|} = \frac{\frac{3}{4}\hat j}{\frac{3}{4}} = \hat j

Why?

The unit vector in the direction of $\vec a$ is $\frac{\vec a}{|\vec a|}$ ⁠: the vector divided by its length.

Since $|\hat j| = 1$ ⁠, the length of $\frac{3}{4}\hat j$ is $\frac{3}{4}$ ⁠.

\begin{aligned} \left|\tfrac{3}{4}\hat j\right| &= \tfrac{3}{4}\,|\hat j| = \tfrac{3}{4}(1) = \tfrac{3}{4} \\ \frac{\frac{3}{4}\hat j}{\frac{3}{4}} &= \left(\tfrac{3}{4} \div \tfrac{3}{4}\right)\hat j \\ &= 1\,\hat j = \hat j \end{aligned}

Watch out: $-\hat j$ is the unit vector in the same direction as the given vector, not the opposite one.

$\therefore$ the required unit vector is $\hat j$ ⁠.

Where this comes from

Chapter 10, Vector Algebra: Unit vector in a given direction.

Question type: A unit vector, or a vector of a given length

How to spot it: click to see

Type 1. A unit vector, or a vector of a given length

Spot it: "unit vector in the direction of", or "a vector of magnitude 8 in the direction of".

Here: The opposite of $-\frac34\hat j$ is $\frac34\hat j$ ⁠. Divide it by its length $\frac34$ to get $\hat j$ ⁠.

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Vector Algebra in past papers: question 45 of 48 · see them all

Q10: Unit vector opposite to −34j^-\frac{3}{4}\hat j−43​j^​