Let vec{c} be the projection vector of vec{b} | vedclass

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(A) The projection vector $\vec{c}$ of $\vec{b}$ on $\vec{a}$ is given by $\vec{c} = \left( \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \right) \vec{a}$

Vedclass · Accepted Answer

$16$

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(A) The projection vector $\vec{c}$ of $\vec{b}$ on $\vec{a}$ is given by $\vec{c} = \left( \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} ight) \vec{a}$.Given $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$,so $|\vec{a}|^2 = 1^2 + 2^2 + 2^2 = 9$.$\vec{b} \cdot \vec{a} = (\lambda \hat{i} + 4\hat{k}) \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) = \lambda + 8$.Thus,$\vec{c} = \frac{\lambda + 8}{9} (\hat{i} + 2\hat{j} + 2\hat{k})$.Given $|\vec{a} + \vec{c}| = 7$. Since $\vec{c}$ is parallel to $\vec{a}$,let $\vec{c} = k\vec{a}$,where $k = \frac{\lambda + 8}{9}$.Then $|\vec{a} + k\vec{a}| = |(1+k)\vec{a}| = |1+k| |\vec{a}| = |1+k| \cdot 3 = 7$.$|1+k| = \frac{7}{3} \Rightarrow 1+k = \frac{7}{3}$ (since $\lambda > 0, k > 0$).$k = \frac{4}{3} \Rightarrow \frac{\lambda + 8}{9} = \frac{4}{3} \Rightarrow \lambda + 8 = 12 \Rightarrow \lambda = 4$.Now,$\vec{b} = 4\hat{i} + 4\hat{k}$ and $\vec{c} = \frac{4}{3}(\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{4}{3}\hat{i} + \frac{8}{3}\hat{j} + \frac{8}{3}\hat{k}$.The area of the parallelogram formed by $\vec{b}$ and $\vec{c}$ is $|\vec{b} 	imes \vec{c}|$.Since $\vec{c}$ is the projection of $\vec{b}$ on $\vec{a}$,$\vec{c} = k\vec{a}$.$|\vec{b} 	imes \vec{c}| = |\vec{b} 	imes k\vec{a}| = |k| |\vec{b} 	imes \vec{a}|$.$\vec{b} 	imes \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 0 & 4 \ 1 & 2 & 2 \end{vmatrix} = \hat{i}(0-8) - \hat{j}(8-4) + \hat{k}(8-0) = -8\hat{i} - 4\hat{j} + 8\hat{k}$.$|\vec{b} 	imes \vec{a}| = \sqrt{(-8)^2 + (-4)^2 + 8^2} = \sqrt{64 + 16 + 64} = \sqrt{144} = 12$.Area $= |k| \cdot 12 = \frac{4}{3} \cdot 12 = 16$.

Let $\vec{c}$ be the projection vector of $\vec{b}=\lambda \hat{i}+4 \hat{k}, \lambda>0$,on the vector $\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k}$. If $|\vec{a}+\vec{c}|=7$,then the area of the parallelogram formed by the vectors $\vec{b}$ and $\vec{c}$ is . . . . . . .

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