The coordinates of the foot of the perpendicular from the point (1, -2, 1) on the plane containing the lines, {{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8} and {{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7}, is :

Question

Accepted Answer

Answer: C. (0, 0, 0) \overrightarrow n = \overrightarrow {{n_1}} 	imes \overrightarrow {{n_2}} = \left| {\begin{matrix} {\widehat i} & {\widehat j} & {\widehat k} \ 6 & 7 & 8 \ 3 & 5 & 7 \ \end{matrix} } ight| = (9, - 18, 9) = (1, -2, 1) 	herefore Equation of plane is 1(x + 1) - 2(y - 1) + (z - 3) = 0 \Rightarrow x - 2y + z = 0 foot to z {{x - 1} \over 1} = {{y + 2} \over { - 2}} = {{z - 1} \over 1} = - {{\left[ {1 + 4 + 1} ight]} \over 6} x = 0, y = 0, z = 0

The coordinates of the foot of the perpendicular from the point (1, $$-$$2, 1) on the plane containing the lines, $${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$$ and $${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7},$$ is :

Answer

Explanation

More JEE Math questions