Question

$A=(x_1,y_1{z}_1)$ and $B=(x_2,y_{2,}{z}_2)$ are two points. lf $l,m,n$ are the d.cs of $CD$ and $l(x_2-x_1)+m({\gamma }_2-y1)+n(z_2-z_1)=0$, then the cosine of the angle between the lines $AB$ and $CD$ is:

• A. ${90}^o$
• B. ${60}^o$
• C. ${30}^o$
• D. ${\frac{1}{2}}^o$

Answer 1

Answer: (A)

${90}^o$

Direction cosines of the line $CD$ are $(l,m,n)$ and thus direction ratios of the line $CD$ are $(l,m,n)$
Now d.r. s of the line $AB$ are $\left(\right(x_2-x_1),(y_2-y_1),(z_2-z_1\left)\right)$
Since, $l(x_2-x_1)+m(y_2-y_1)+n(z_2-z_1)=0$
Thererfore, $\cos \theta =0$
$\Rightarrow \theta =\frac{\pi }{2}$
Ans: A