Question 36

Mathematics Scalar and Vector Products Hard

Let \(\vec{a}=2\widehat{i}\, +\widehat{j}\, +2\widehat{k}\) , \(\vec{b}=\widehat{i}-\widehat{j}+2\widehat{k}\) and \(\vec{c}=\widehat{i}+\widehat{j}-2\widehat{k}\) are are three vectors. Then, a vector in the plane of \(\vec{a}\) and \(\vec{c}\) whose projection on \(\vec{b}\) is of magnitude \(\frac{1}{\sqrt{6}}\) is

(A) \(3\widehat{i}\, -2\widehat{j}\)
(B) \(3\widehat{i}\, +2\widehat{j}\)
(C) \(\begin{array}{ll}{2\widehat{i}}+3\widehat{j}-\widehat{k}\end{array}\)
(D) \(\begin{array}{ll}{3\widehat{i}}+2\widehat{j}+\widehat{k}\end{array}\)
✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Any vector that is coplanar with \(\vec{a}\) and \(\vec{c}\) is: \[ \vec{a} + \lambda \vec{c} = (2 + \lambda)\hat{i} + (1 + \lambda)\hat{j} + (2 - 2\lambda)\hat{k} \] The projection on \(\vec{b}\) is: \[ (\vec{a} + \lambda \vec{c}) \cdot \hat{b} = \frac{(2 + \lambda) - (1 + \lambda) + 2(2 - 2\lambda)}{\sqrt{1^2 + (-1)^2 + 2^2}} = \frac{5 - 4\lambda}{\sqrt{6}} \] Given that the projection on \(\vec{b}\) is of magnitude \(\frac{1}{\sqrt{6}}\): \[ 5 - 4\lambda = 1 \implies \lambda = 1 \] The vector is: \[ \vec{a} + \lambda \vec{c} = (2 + \lambda)\hat{i} + (1 + \lambda)\hat{j} + (2 - 2\lambda)\hat{k} = 3\hat{i} + 2\hat{j} \]

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