Question

The point of intersection of the line joining the points $(-3,4,-8)$ and $(5,-6,4)$ with the $XY$-plane is

• A. $(\frac{7}{3},-\frac{8}{3},0)$
• B. $(-\frac{7}{3},-\frac{8}{3},0)$
• C. $(-\frac{7}{3},\frac{8}{3},0)$
• D. $(\frac{7}{3},\frac{8}{3},0)$

Answer 1

The correct option is A $(\frac{7}{3},-\frac{8}{3},0)$

$\overrightarrow{r}=\overrightarrow{a}+x(\overrightarrow{b}-\overrightarrow{a})\overrightarrow{a}=-3\hat{i}+4\hat{j}+(-8)\hat{k}\overrightarrow{b}=5\hat{i}+(-6)\hat{j}+4\hat{k}\overrightarrow{r}=-3\hat{i}+4\hat{j}+(-8)\hat{k}+x(8\hat{i}+(-10)\hat{j}+12\hat{k})=(8x-3)\hat{i}+(4-10x)\hat{j}+(12x-8)\hat{k}$

On $XY$ plane $\hat{k}$ will be $0$

So $12x-8=0$ it means $x=\frac{2}{3}$

Putting the value of $x$ in component of $\hat{i}$ and $\hat{j}$ we get

$8\times \frac{2}{3}-3$ and $4-10\times \frac{2}{3}$

$\frac{7}{3}and\frac{-8}{3}$

So correct optiopn will be option A.