JEE Advanced 2013 · Paper 1 · Mathematics Q49

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MathematicsSingle correct+2Question 49 of 60

Perpendiculars are drawn from points on the line $\dfrac{x+2}2=\dfrac{y+1}{-1}=\dfrac z3$ to the plane $x+y+z=3$ . The feet of perpendiculars lie on the line

\dfrac x5=\dfrac{y-1}8=\dfrac{z-2}{-13}

\dfrac x2=\dfrac{y-1}3=\dfrac{z-2}{-5}

\dfrac x4=\dfrac{y-1}3=\dfrac{z-2}{-7}

\dfrac x2=\dfrac{y-1}{-7}=\dfrac{z-2}5

Answer and worked solution

Official answer: D

Foot of the perpendicular. A point of the line is $(-2+2t,\,-1-t,\,3t)$ , with $x+y+z-3=4t-6$ . Subtract $\frac{4t-6}3(1,1,1)$ :
$\left(\tfrac{2t}3,\ 1-\tfrac{7t}3,\ 2+\tfrac{5t}3\right)=(0,1,2)+\tfrac t3(2,-7,5).$

Line: $\dfrac x2=\dfrac{y-1}{-7}=\dfrac{z-2}5$ .

(D)

Topics:Equation of a line in space Angles and distances

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JEE Advanced 2013 · Paper 1, Mathematics question 49