If the components of vec{a}=alpha hat{i}+beta | vedclass

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(D) Let $\vec{a}_{\parallel}$ be the component of $\vec{a}$ along $\vec{b}$ and $\vec{a}_{\perp}$ be the component of $\vec{a}$ perpendicular to $\vec

Vedclass · Accepted Answer

$26$

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(D) Let $\vec{a}_{\parallel}$ be the component of $\vec{a}$ along $\vec{b}$ and $\vec{a}_{\perp}$ be the component of $\vec{a}$ perpendicular to $\vec{b}$.Given $\vec{a}_{\parallel} = \frac{16}{11}(3 \hat{i} + \hat{j} - \hat{k})$ and $\vec{a}_{\perp} = \frac{1}{11}(-4 \hat{i} - 5 \hat{j} - 17 \hat{k})$.Since $\vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp}$,we have:$\vec{a} = \frac{16}{11}(3 \hat{i} + \hat{j} - \hat{k}) + \frac{1}{11}(-4 \hat{i} - 5 \hat{j} - 17 \hat{k})$$\vec{a} = \frac{48-4}{11} \hat{i} + \frac{16-5}{11} \hat{j} + \frac{-16-17}{11} \hat{k}$$\vec{a} = \frac{44}{11} \hat{i} + \frac{11}{11} \hat{j} - \frac{33}{11} \hat{k}$$\vec{a} = 4 \hat{i} + \hat{j} - 3 \hat{k}$Comparing with $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,we get $\alpha = 4$,$\beta = 1$,and $\gamma = -3$.Therefore,$\alpha^2 + \beta^2 + \gamma^2 = (4)^2 + (1)^2 + (-3)^2 = 16 + 1 + 9 = 26$.

If the components of $\vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ respectively,are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})$,then $\alpha^2+\beta^2+\gamma^2$ is equal to :

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