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Let a⃗=2i^−j^,b⃗=−4j^+k and c⃗=i^+2k^\vec{a} = 2\hat{i} - \hat{j}, \vec{b} =- 4\hat{j} + k\,\text{and}\,\vec{c} = \hat{i} + 2\hat{k}. If d⃗\vec{d} is a vector perpendicular to both a⃗\vec{a} and b⃗\vec{b} such that c⃗⋅d⃗=34\vec{c} \cdot \vec{d} = 34, then ∣d⃗∣|\vec{d}| is equal to

Solution

✅ Correct Option: 2

We're given:

a⃗=2i^−j^\vec{a} = 2\hat{i} - \hat{j}, b⃗=−4j^+k^\vec{b} = -4\hat{j} + \hat{k}, c⃗=i^+2k^\vec{c} = \hat{i} + 2\hat{k}

Since d⃗\vec{d} is perpendicular to both a⃗\vec{a} and b⃗\vec{b}, it must be parallel to their cross product.

So, d⃗=λ(a⃗×b⃗)\vec{d} = \lambda(\vec{a} \times \vec{b}) for some scalar λ\lambda.


Writing the vectors with all three components:

a⃗=2i^−1j^+0k^\vec{a} = 2\hat{i} - 1\hat{j} + 0\hat{k}, b⃗=0i^−4j^+1k^\vec{b} = 0\hat{i} - 4\hat{j} + 1\hat{k}

a⃗×b⃗=∣i^j^k^2−100−41∣\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 0 \\ 0 & -4 & 1 \end{vmatrix}

=i^[(−1)(1)−(0)(−4)]−j^[(2)(1)−(0)(0)]+k^[(2)(−4)−(−1)(0)]= \hat{i}[(-1)(1) - (0)(-4)] - \hat{j}[(2)(1) - (0)(0)] + \hat{k}[(2)(-4) - (-1)(0)]

=−i^−2j^−8k^= -\hat{i} - 2\hat{j} - 8\hat{k}

So, d⃗=λ(−i^−2j^−8k^)\vec{d} = \lambda(-\hat{i} - 2\hat{j} - 8\hat{k})


Using the condition c⃗⋅d⃗=34\vec{c} \cdot \vec{d} = 34:

c⃗⋅d⃗=(i^+2k^)⋅λ(−i^−2j^−8k^)\vec{c} \cdot \vec{d} = (\hat{i} + 2\hat{k}) \cdot \lambda(-\hat{i} - 2\hat{j} - 8\hat{k})

=λ[(−1)+(0)+(−16)]= \lambda[(-1) + (0) + (-16)]

=−17λ= -17\lambda

Setting this equal to 3434:

−17λ=34-17\lambda = 34

λ=−2\lambda = -2


d⃗=−2(−i^−2j^−8k^)=2i^+4j^+16k^\vec{d} = -2(-\hat{i} - 2\hat{j} - 8\hat{k}) = 2\hat{i} + 4\hat{j} + 16\hat{k}

∣d⃗∣=22+42+162|\vec{d}| = \sqrt{2^2 + 4^2 + 16^2}

=4+16+256= \sqrt{4 + 16 + 256}

=276= \sqrt{276}

=269= 2\sqrt{69}


∣d⃗∣=269|\vec{d}| = 2\sqrt{69}

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