JEE Main 2017 · 8 Apr · Shift 1 · Mathematics Q82

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MathematicsSingle correct+4 / -1Question 82 of 90

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

(2, −4, 2)

(−1, 2, −1)

(0, 0, 0)

(1, 1, 1)

Answer and worked solution

Official answer: C

The normal is $(6,7,8)\times(3,5,7)=(9,-18,9)\parallel(1,-2,1)$ . Through $(-1,1,3)$ the plane is $x-2y+z=0$ . For (1, −2, 1), $x-2y+z=6$ , so the foot is
$(1,-2,1)-\frac66(1,-2,1)=(0,0,0).$

Topics:Equation of a plane Angles and distances

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JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 82