Question
Let $\vec{\text{a }}=\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}},\vec{\text{b }}=3\hat{\text{i}}-2\hat{\text{j}}+7\hat{\text{k}}\text{ and }\vec{\text{ c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}.$ Find a vector $\vec{\text{ p}}$  which is perpendicular to both $\vec{\text{a }}$ and $\vec{\text{ b}}$ and $\vec{\text{ p}}\cdot\vec{\text{ c}}$ = 18.

Answer

$\vec{\text{p}}\text{ is }\bot\text{ }\text{ to both }\vec{\text{a}}\text{ and }\vec{\text{b}}\Rightarrow\vec{\text{p}}=\lambda\text{ }(\vec{\text{a}}\times\vec{\text{b}})$

Now $\vec{\text{a}}\times\vec{\text{b}}$ $\begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix}$ =
$32\hat{\text{i}}-\hat{\text{j}}-14\hat{\text{k}}$

Given that $\vec{\text{ p}}\cdot\vec{\text{ c}}$ = 18 $\Rightarrow\lambda(32\hat{\text{i}}-\hat{\text{j}}-14\hat{\text{k}})\cdot(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})=18$

OR $\lambda(64+1-56)=18\Rightarrow\lambda=2$

$\therefore\vec{\text{p}}=64\hat{\text{i}}-2\hat{\text{j}}-28\hat{\text{k}}$.

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