CUET Mathematics 2024 - The unit vector perpendicular to each of the vectors a+b and a-b, where a=i+j+k and b=i+2 j+3 k, is : | PYQs + Solutions

CUET Mathematics 2024

Algebra

Vector Algebra

Medium

The unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ , where $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}$ , is :

\frac{1}{\sqrt{6}} \hat{i}+\frac{2}{\sqrt{6}} \hat{\mathrm{j}}+\frac{1}{\sqrt{6}} \hat{\mathrm{k}}

-\frac{1}{\sqrt{6}} \hat{\mathrm{i}}+\frac{1}{\sqrt{6}} \hat{\mathrm{j}}-\frac{1}{\sqrt{6}} \hat{\mathrm{k}}

-\frac{1}{\sqrt{6}} \hat{\mathrm{i}}+\frac{2}{\sqrt{6}} \hat{\mathrm{j}}+\frac{2}{\sqrt{6}} \hat{\mathrm{k}}

-\frac{1}{\sqrt{6}} \hat{\mathrm{i}}+\frac{2}{\sqrt{6}} \hat{\mathrm{j}}-\frac{1}{\sqrt{6}} \hat{\mathrm{k}}

✅ Correct Option: 4

To find the unit vector perpendicular to both

\vec{a}+\vec{b}

and

\vec{a}-\vec{b}

, we need to calculate their cross product and then normalize it.

Given:

\newline

\vec{a} = \hat{i} + \hat{j} + \hat{k}

\newline

\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}

Let's calculate

\vec{a}+\vec{b}

and

\vec{a}-\vec{b}

\vec{a} + \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k}) = 2\hat{i} + 3\hat{j} + 4\hat{k}

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

A vector perpendicular to both would be their cross product:

\newline

(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})

Computing this cross product using the determinant:

\newline

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix}

= (3 \times (-2) - 4 \times (-1))\hat{i} - (2 \times (-2) - 4 \times 0)\hat{j} + (2 \times (-1) - 3 \times 0)\hat{k}

= (-6 + 4)\hat{i} - (-4)\hat{j} + (-2)\hat{k}

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

To make this a unit vector, we divide by its magnitude:

Magnitude =

\sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}

Therefore, the unit vector is:

\dfrac{1}{2\sqrt{6}}(-2\hat{i} + 4\hat{j} - 2\hat{k}) = \dfrac{1}{\sqrt{6}}(-\hat{i} + 2\hat{j} - \hat{k})