CUET UG 2025 — Mathematics Algebra

A vector perpendicular to both given vectors can be found using the cross product.

The given vectors are:

$$\vec{a} = 4\hat{i} - \hat{j} + 8\hat{k}$$a=4i^−j^​+8k^

$$\vec{b} = 0\hat{i} - \hat{j} + \hat{k}$$b=0i^−j^​+k^

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The cross product $\vec{a} \times \vec{b}$a×b is:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 8 \\ 0 & -1 & 1 \end{vmatrix}$$a×b=​i^40​j^​−1−1​k^81​​

$$= \hat{i}\begin{vmatrix} -1 & 8 \\ -1 & 1 \end{vmatrix} - \hat{j}\begin{vmatrix} 4 & 8 \\ 0 & 1 \end{vmatrix} + \hat{k}\begin{vmatrix} 4 & -1 \\ 0 & -1 \end{vmatrix}$$=i^​−1−1​81​​−j^​​40​81​​+k^​40​−1−1​​

$$= \hat{i}[(-1)(1) - (8)(-1)] - \hat{j}[(4)(1) - (8)(0)] + \hat{k}[(4)(-1) - (-1)(0)]$$=i^[(−1)(1)−(8)(−1)]−j^​[(4)(1)−(8)(0)]+k^[(4)(−1)−(−1)(0)]

$$= \hat{i}[-1 + 8] - \hat{j}[4 - 0] + \hat{k}[-4 - 0]$$=i^[−1+8]−j^​[4−0]+k^[−4−0]

$$= 7\hat{i} - 4\hat{j} - 4\hat{k}$$=7i^−4j^​−4k^

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The magnitude of this vector is:

$$|\vec{v}| = \sqrt{(7)^2 + (-4)^2 + (-4)^2}$$∣v∣=(7)2+(−4)2+(−4)2​

$$= \sqrt{49 + 16 + 16}$$=49+16+16​

$$= \sqrt{81}$$=81​

$$= 9$$=9

The vector $7\hat{i} - 4\hat{j} - 4\hat{k}$7i^−4j^​−4k^ has magnitude 9 and is perpendicular to both given vectors.