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  4. If a line passing through (- 2 , 1 , α ) and (4,1 , 2) is perpendicular to the vector 3 hati-4 hatj+5 hatk and parallel to the plane containing vectors hati+2β hatk and 2β hatj+α hatk(∀ β ≠ 0) , then 10(α + β ) is equal to

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Q. If a line passing through $\left(- 2 , 1 , \alpha \right)$ and $\left(4,1 , 2\right)$ is perpendicular to the vector $3\hat{i}-4\hat{j}+5\hat{k}$ and parallel to the plane containing vectors $\hat{i}+2\beta \hat{k}$ and $2\beta \hat{j}+\alpha \hat{k}\left(\forall \beta \neq 0\right)$ , then $10\left(\alpha + \beta \right)$ is equal to

NTA AbhyasNTA Abhyas 2020Vector Algebra

Solution:

Let, coordinates of $A,B$ are respectively $\left(- 2,1 , \alpha \right),\left(4,1 , 2\right)$ respectively
$\overset{ \rightarrow }{A B}=6\hat{i}+\left(2 - \alpha \right)\hat{k}$
$\because \overset{ \rightarrow }{A B}$ is perpendicular to $3\hat{i}-4\hat{j}+5\hat{k}$
$\Rightarrow 18+5\left(2 - \alpha \right)=0\Rightarrow \alpha =\frac{28}{5}$
Also, $\overset{ \rightarrow }{A B},\hat{i}+2\beta \hat{k},2\beta \hat{j}+\alpha \hat{k}$ are coplanar
$\Rightarrow \begin{vmatrix} 6 & 0 & 2-\alpha \\ 1 & 0 & 2\beta \\ 0 & 2\beta & \alpha \end{vmatrix}=0$
$\Rightarrow 2\beta \left(12 \beta + \alpha - 2\right)=0\Rightarrow \beta =0$ or $12\beta =2-\alpha $
$12\beta =2-\frac{28}{5}$
$12\beta =-\frac{18}{5}\Rightarrow \beta =-\frac{3}{10}$