If the points with position vectors i+10j+13k,6i+11j+11k,(9)/(2)i+ j-8k are collinear, the…

Question

If the points with position vectors $\alpha \hat{i}+10\hat{j}+13\hat{k},6\hat{i}+11\hat{j}+11\hat{k},\frac{9}{2}\hat{i}+\beta \hat{j}-8\hat{k}$ are collinear, then $(19\alpha -6\beta {)}^{2}$ is equal to :

Accepted Answer

Given : Points with position vectors

$\alpha \hat{i}+10\hat{j}+13\hat{k},6\hat{i}+11\hat{j}+11\hat{k}$

and $\frac{9}{2}\hat{i}+\beta \hat{j}-8\hat{k}$ are collinear.

So, $\frac{\alpha -6}{6-\frac{9}{2}}=\frac{10-11}{11-\beta }=\frac{13-11}{11+8}$

$\begin{gathered} \Rightarrow \frac{2(\alpha -6)}{3}=\frac{-1}{11-\beta }=\frac{2}{19} \ \Rightarrow \frac{2}{3}(\alpha -6)=\frac{2}{19} \end{gathered}$

$\begin{gathered} \Rightarrow 19\alpha -114=3\Rightarrow 19\alpha =117 \ \Rightarrow \alpha =\frac{117}{19} \end{gathered}$

And, $\frac{-1}{11-\beta }=\frac{2}{19}$

$\begin{gathered} \Rightarrow -19=22-2\beta \ \Rightarrow 2\beta =41 \ \Rightarrow \beta =\frac{41}{2} \end{gathered}$

$\begin{gathered} 	herefore (19\alpha -6\beta {)}^{2}={(19	imes \frac{117}{19}-\frac{6	imes 41}{2})}^{2} \ =(117-123{)}^{2}=36 \end{gathered}$

Solutions · 1