Let vec{a}=alpha hat{i}+hat{j}+beta hat{k} an | vedclass

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(D) Given $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$.The cross product is $\vec{a} 	imes \vec{b} = \b

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$\frac{46}{5}$

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(D) Given $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$.The cross product is $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \alpha & 1 & \beta \ 3 & -5 & 4 \end{vmatrix} = (4+5\beta)\hat{i} + (3\beta-4\alpha)\hat{j} + (-5\alpha-3)\hat{k}$.Comparing this with $-\hat{i}+9\hat{j}+12\hat{k}$,we get:$4+5\beta = -1 \Rightarrow 5\beta = -5 \Rightarrow \beta = -1$.$-5\alpha-3 = 12 \Rightarrow -5\alpha = 15 \Rightarrow \alpha = -3$.Check: $3\beta-4\alpha = 3(-1)-4(-3) = -3+12 = 9$ (Correct).So,$\vec{a} = -3\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} + 4\hat{k}$.Then $\vec{b}-2\vec{a} = (3 - 2(-3))\hat{i} + (-5 - 2(1))\hat{j} + (4 - 2(-1))\hat{k} = 9\hat{i} - 7\hat{j} + 6\hat{k}$.And $\vec{b}+\vec{a} = (3-3)\hat{i} + (-5+1)\hat{j} + (4-1)\hat{k} = -4\hat{j} + 3\hat{k}$.The projection of $\vec{v}_1 = \vec{b}-2\vec{a}$ on $\vec{v}_2 = \vec{b}+\vec{a}$ is $\frac{\vec{v}_1 \cdot \vec{v}_2}{|\vec{v}_2|}$.$\vec{v}_1 \cdot \vec{v}_2 = (9)(0) + (-7)(-4) + (6)(3) = 0 + 28 + 18 = 46$.$|\vec{v}_2| = \sqrt{0^2 + (-4)^2 + 3^2} = \sqrt{16+9} = 5$.Projection = $\frac{46}{5}$.

Let $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$ be two vectors,such that $\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}$. Then the projection of $\vec{b}-2 \vec{a}$ on $\vec{b}+\vec{a}$ is equal to.

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