CUET UG 2025 — Mathematics Arithmetic

If $\vec{d}$d is perpendicular to both $\vec{a}$a and $\vec{b}$b, then $\vec{d}$d must be parallel to $\vec{a} \times \vec{b}$a×b, so:

$$\vec{d} = \lambda(\vec{a} \times \vec{b})$$d=λ(a×b)

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$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix}$$a×b=​i^13​j^​4−2​k^27​​

$$= \hat{i}(28 + 4) - \hat{j}(7 - 6) + \hat{k}(-2 - 12)$$=i^(28+4)−j^​(7−6)+k^(−2−12)

$$\vec{a} \times \vec{b} = 32\hat{i} - \hat{j} - 14\hat{k}$$a×b=32i^−j^​−14k^

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$$\vec{d} = \lambda(32\hat{i} - \hat{j} - 14\hat{k})$$d=λ(32i^−j^​−14k^)

Using $\vec{c} \cdot \vec{d} = 14$c⋅d=14 with $\vec{c} = 2\hat{i} + \hat{j} + 4\hat{k}$c=2i^+j^​+4k^:

$$\vec{c} \cdot \vec{d} = \lambda\big[(2)(32) + (1)(-1) + (4)(-14)\big]$$c⋅d=λ[(2)(32)+(1)(−1)+(4)(−14)]

$$= \lambda(64 - 1 - 56)$$=λ(64−1−56)

$$= 7\lambda$$=7λ

$$7\lambda = 14$$7λ=14

$$\lambda = 2$$λ=2

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$$\vec{d} = 2(32\hat{i} - \hat{j} - 14\hat{k})$$d=2(32i^−j^​−14k^)

$$\vec{d} = 64\hat{i} - 2\hat{j} - 28\hat{k}$$d=64i^−2j^​−28k^