Let vec{a} = 2hat{i} - hat{j} + hat{k},vec{b} | vedclass

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(B) Let $\vec{d} = \vec{b} + \lambda \vec{c}$.Then $\vec{d} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(\hat{i} + \hat{j} - 2\hat{k}) = (1+\lambda)\hat

Vedclass · Accepted Answer

$2\hat{i} + 3\hat{j} - 3\hat{k}$

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(B) Let $\vec{d} = \vec{b} + \lambda \vec{c}$.Then $\vec{d} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(\hat{i} + \hat{j} - 2\hat{k}) = (1+\lambda)\hat{i} + (2+\lambda)\hat{j} - (1+2\lambda)\hat{k}$.The projection of $\vec{d}$ on $\vec{a}$ is given by $\frac{|\vec{d} \cdot \vec{a}|}{|\vec{a}|} = \sqrt{\frac{2}{3}}$.First,calculate $\vec{d} \cdot \vec{a} = 2(1+\lambda) - 1(2+\lambda) + 1(-1-2\lambda) = 2 + 2\lambda - 2 - \lambda - 1 - 2\lambda = -\lambda - 1$.Also,$|\vec{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.So,$\frac{|-\lambda - 1|}{\sqrt{6}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{2}}{\sqrt{3} \cdot \sqrt{2}} = \frac{2}{\sqrt{6}}$.Thus,$|-\lambda - 1| = 2$,which means $|\lambda + 1| = 2$.This gives $\lambda + 1 = 2$ or $\lambda + 1 = -2$,so $\lambda = 1$ or $\lambda = -3$.For $\lambda = 1$,$\vec{d} = (1+1)\hat{i} + (2+1)\hat{j} - (1+2)\hat{k} = 2\hat{i} + 3\hat{j} - 3\hat{k}$.For $\lambda = -3$,$\vec{d} = (1-3)\hat{i} + (2-3)\hat{j} - (1-6)\hat{k} = -2\hat{i} - \hat{j} + 5\hat{k}$.Comparing with the options,the correct vector is $2\hat{i} + 3\hat{j} - 3\hat{k}$.

Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} - 2\hat{k}$ be three vectors. $A$ vector of the type $\vec{b} + \lambda \vec{c}$ for some scalar $\lambda$,whose projection on $\vec{a}$ is of magnitude $\sqrt{\frac{2}{3}}$ is

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