Question

The point of intersection $\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}$ and $\frac{x+3}{\left(-36\right)}=\frac{y-3}{2}=\frac{z-6}{4}$ is

• A. $(5,7,-2)$
• B. $(-3,3,6)$
• C. $(2,10,4)$
• D. $(21,\frac{5}{3},\frac{10}{3})$

Answer 1

The correct option is D

$(21,\frac{5}{3},\frac{10}{3})$

The explanation for the correct answer.

Solve for the point of intersection.

The given equation of lines are

$\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}=k...\left(1\right)$

and $\frac{x+3}{\left(-36\right)}=\frac{y-3}{2}=\frac{z-6}{4}...\left(2\right)$

Any point on the line $\left(1\right)$ is $(3k+5,7-k,k-2).$

It should lie on line $\left(2\right)$

$\Rightarrow \frac{3k+5+3}{-36}=\frac{7-k-3}{2}=\frac{k-2-6}{4}$

Now, to find the value of $k$,

$$\begin{gathered} \frac{7-k-3}{2}=\frac{k-2-6}{4} \\ \Rightarrow \frac{4-k}{2}=\frac{k-8}{4} \\ \Rightarrow 16-4k=2k-16 \\ \Rightarrow 6k=32 \\ \Rightarrow k=\frac{32}{6} \\ \Rightarrow k=\frac{16}{3} \end{gathered}$$

On solving, we get $k=\frac{16}{3}$

The coordinates of the intersection point are$\left(3\left(\frac{16}{3}\right)+5,7-\frac{16}{3},\frac{16}{3}-2\right)=$ $(21,\frac{5}{3},\frac{10}{3})$

Hence, option(D) is the correct answer.