Discussion Forum

1)

If the position vectors of the points A, B, C, D given by $\hat{i}+2\hat{j}+3\hat{k}$ , $2\hat{i}-\hat{j}+2\hat{k}$, 

$\frac{1}{4}(7 \hat{i}+15\hat{j}+15 \hat{k})$ and    $\frac{1}{3}[7\hat{i}+2\hat{j}+(5+3a)\hat{k}]$ respectively are such that |AC| =|BD| , then $16(3a-1)^{2}$=

Answer:

Option D

Explanation:

Given ,

A= ($\hat{i}+2\hat{j}+3\hat{k}$ )

B= ( $2\hat{i}-\hat{j}+2\hat{k}$)

C= $\frac{1}{4}(7 \hat{i}+15\hat{j}+15 \hat{k})$

D= $\frac{1}{3}[7\hat{i}+2\hat{j}+(5+3a)\hat{k}]$

$AC=\left(\frac{7}{4}\hat{i}+\frac{15}{4}\hat{j}+\frac{15}{4}\hat{k}\right)-(\hat{i}+2 \hat{j}+3\hat{k})$

 AC= $\frac{3}{4} \hat{i}+\frac{7}{4} \hat{j}+\frac{3}{4} \hat{k}$

 and BD= $ \left(\frac{7}{3}\hat{i}+\frac{2}{3}\hat{j}+\left(\frac{5+3a}{3}\right)\hat{k}\right)-(2\hat{i}-\hat{j}+2\hat{k})$

  $BD=\frac{1}{3}\hat{i}+\frac{5}{3}\hat{j}+\left(\frac{3a-1}{3}\right)\hat{k}$ 

 Also given,

|AC|=|BD|

 $\because$    $\left(\frac{3}{4}\right)^{2}+\left(\frac{7}{4}\right)^{2}+\left(\frac{3}{4}\right)^{2}=\left(\frac{1}{3}\right)^{2}+\left(\frac{5}{3}\right)^{2}+\left(\frac{3a-1}{3}\right)^{2}$

 $\Rightarrow$    $\frac{9+49+9}{16}=\frac{1+25+(3a-1)^{2}}{9}$

 $\Rightarrow$      $\frac{67}{16}= \frac{26+(3a-1)^{2}}{9}$

 $\Rightarrow$     603=416+16(3a-1)²

$\Rightarrow$       $16(3a-1)^{2}=187$