Board 2023 Class 12 Maths Q22 Solution: Vectors of magnitude 3√3 collinear to i + j + k

The question

Find all the vectors of magnitude $3\sqrt3$ which are collinear to vector $\hat i + \hat j + \hat k$ ⁠.

Exam answer

Let $\vec a = \hat i + \hat j + \hat k$ ⁠. Then $|\vec a| = \sqrt{1 + 1 + 1} = \sqrt3$ ⁠.

Why?

The magnitude of a vector comes from its components:

\begin{aligned} &|x\hat i + y\hat j + z\hat k| \\ &= \sqrt{x^2 + y^2 + z^2} \end{aligned}

Here each component is 1, so each square is 1.

\begin{aligned} |\vec a| &= \sqrt{1^2 + 1^2 + 1^2} \\ &= \sqrt{1 + 1 + 1} = \sqrt3 \end{aligned}

Unit vector along $\vec a$ ⁠:

\hat a = \frac{\hat i}{\sqrt3} + \frac{\hat j}{\sqrt3} + \frac{\hat k}{\sqrt3}

Why?

Dividing a vector by its magnitude gives the unit vector in the same direction:

\hat a = \frac{\vec a}{|\vec a|} = \frac{\hat i + \hat j + \hat k}{\sqrt3}

The vectors of magnitude $3\sqrt3$ collinear to $\vec a$ are $\pm 3\sqrt3\,\hat a$ ⁠:

\pm 3\sqrt3\left(\frac{\hat i + \hat j + \hat k}{\sqrt3}\right)

Why?

Collinear vectors lie along the same line, so they point either along $\vec a$ or the opposite way.

A vector of magnitude $m$ along $\hat a$ is $m\hat a$ ⁠, and the one pointing the opposite way is $-m\hat a$ ⁠. Here $m = 3\sqrt3$ ⁠.

Watch out: the question asks for all the vectors, so both signs are needed.

= \pm\left(3\hat i + 3\hat j + 3\hat k\right)

Why?

The $\sqrt3$ cancels:

3\sqrt3 \times \frac{1}{\sqrt3} = 3

So each of $\hat i$ ⁠, $\hat j$ ⁠, $\hat k$ gets the number 3:

\begin{aligned} &3\sqrt3\left(\frac{\hat i}{\sqrt3} + \frac{\hat j}{\sqrt3} + \frac{\hat k}{\sqrt3}\right) \\ &= 3\hat i + 3\hat j + 3\hat k \end{aligned}

Check the magnitude:

\begin{aligned} &\sqrt{3^2 + 3^2 + 3^2} \\ &= \sqrt{9 + 9 + 9} = \sqrt{27} \\ &= \sqrt{9 \times 3} = 3\sqrt3 \end{aligned}

$\therefore$ Required vectors are $3\hat i + 3\hat j + 3\hat k$ and $-3\hat i - 3\hat j - 3\hat k$ ⁠.

Where this comes from

Chapter 10, Vector Algebra: Vectors of given magnitude collinear to a vector.

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 length wanted is $3\sqrt3$ and $\hat i + \hat j + \hat k$ has length $\sqrt3$ ⁠. Multiply its unit vector by $\pm 3\sqrt3$ ⁠: collinear allows both directions.

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Q22: Vectors of magnitude 333\sqrt333​ collinear to i^+j^+k^\hat i + \hat j + \hat ki^+j^​+k^

Q22: Vectors of magnitude $3\sqrt3$ collinear to $\hat i + \hat j + \hat k$