CUET Preparation Today
Distances of the point $P(x_2, y_2, z_2)$ from the line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$, where l, m, n are the direction cosines of the line, is |
$\sqrt{l^2(x_2-x)^2 +m^2(y_2-y_1)^2+n^2(z_2-z_1)^2}$ $|l(x_2-x) +m^2(y_2-y_1)+n(z_2-z_1)|$ $\sqrt{(x_2-x)^2 +(y_2-y_1)^2+(z_2-z_1)^2-l(x_2-x)^2 +m(y_2-y_1)^2+n(z_2-z_1)^2}$ none of these |
$\sqrt{(x_2-x)^2 +(y_2-y_1)^2+(z_2-z_1)^2-l(x_2-x)^2 +m(y_2-y_1)^2+n(z_2-z_1)^2}$ |
We have, $QM = Projection of QP on the given line $⇒ QM = |(x_2-x_1)l+(y_2-y_1)m + (z_2-z_1)n|$ and, $PQ= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2 +(z_2-z_1)^2}$
In right angled triangle PMQ, we have $PM = \sqrt{PQ^2-QM^2}$ $⇒ PM =\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2+(z_2-z_1)^2-l(x_2-x_1)^2 +m(y_2-y_1)^2+n(z_2-z_1)^2}$ |
