Let vec{a}=hat{i}+4 hat{j}+2 hat{k}, vec{b}=3 | vedclass

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(A) Let $\vec{d}=d_{1} \hat{i}+d_{2} \hat{j}+d_{3} \hat{k}$.Since $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$,we have $\vec{d} \cdot \v

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$\frac{1}{3}(160 \hat{i}-5 \hat{j}-70 \hat{k})$

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(A) Let $\vec{d}=d_{1} \hat{i}+d_{2} \hat{j}+d_{3} \hat{k}$.Since $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$,we have $\vec{d} \cdot \vec{a} = 0$ and $\vec{d} \cdot \vec{b} = 0$.$\vec{d} \cdot \vec{a} = d_{1} + 4d_{2} + 2d_{3} = 0$ .....$(i)$$\vec{d} \cdot \vec{b} = 3d_{1} - 2d_{2} + 7d_{3} = 0$ .....$(ii)$Also,it is given that $\vec{c} \cdot \vec{d} = 15$,so $2d_{1} - d_{2} + 4d_{3} = 15$ .....$(iii)$From $(i)$,$d_{1} = -4d_{2} - 2d_{3}$. Substituting this into $(ii)$:$3(-4d_{2} - 2d_{3}) - 2d_{2} + 7d_{3} = 0 \Rightarrow -12d_{2} - 6d_{3} - 2d_{2} + 7d_{3} = 0 \Rightarrow -14d_{2} + d_{3} = 0 \Rightarrow d_{3} = 14d_{2}$.Substituting $d_{3} = 14d_{2}$ into $(i)$:$d_{1} + 4d_{2} + 2(14d_{2}) = 0 \Rightarrow d_{1} + 4d_{2} + 28d_{2} = 0 \Rightarrow d_{1} = -32d_{2}$.Now substitute $d_{1}$ and $d_{3}$ into $(iii)$:$2(-32d_{2}) - d_{2} + 4(14d_{2}) = 15 \Rightarrow -64d_{2} - d_{2} + 56d_{2} = 15 \Rightarrow -9d_{2} = 15 \Rightarrow d_{2} = -\frac{15}{9} = -\frac{5}{3}$.Then $d_{1} = -32(-\frac{5}{3}) = \frac{160}{3}$ and $d_{3} = 14(-\frac{5}{3}) = -\frac{70}{3}$.Thus,$\vec{d} = \frac{160}{3} \hat{i} - \frac{5}{3} \hat{j} - \frac{70}{3} \hat{k} = \frac{1}{3}(160 \hat{i} - 5 \hat{j} - 70 \hat{k})$.

Let $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k} .$ Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b},$ and $\vec{c} \cdot \vec{d}=15.$

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