Question

The ratio in which the plane $\overrightarrow{r}.(\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k})=17$ divides the line joining the points $-2\overrightarrow{i}+4\overrightarrow{j}+7\overrightarrow{k}$ and $3\overrightarrow{i}+5\overrightarrow{j}+8\overrightarrow{k}$

• A. 1:5
• B. 1:10`
• C. 3:5
• D. 3:10

Answer 1

The correct option is A 1:5

Given , $\overline{r}.(\hat{i}-2\hat{j}+3\hat{k})=17$ divides the line joining the points $=-2\hat{i}+4\hat{j}+7\hat{k}$ and $3\hat{i}+5\hat{j}+8\hat{k}$

We know that, if plane divides line in m : n ratio , then point of contact $=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n})$

The plane in cartesian form is $x-2y+3z-17=0...eq1$

Points are (-2, 4, 7) and (3, 5, 8)

Let the plane divides joining line at A in P : 1 ratio.

$\Rightarrow A=[\frac{3p-2}{p+1},\frac{5p+4}{p+1},\frac{8p+7}{p+1}]$

As A lies on plane 1, we get

$\frac{3p-2}{p+1}-2\left(\frac{5p+4}{p+1}\right)+3\left(\frac{8p+7}{p+1}\right)=17\Rightarrow 3p-2-10p-8+24p+21=17p+17\Rightarrow 17p+11=17p+17$

which is not possible.

Hence , the values in question are wrong.