JEE Main 2016 · 3 Apr · Shift 1 · Mathematics Q83

MathematicsSingle correct+4 / -1Question 83 of 90

The distance of the point (1, −5, 9) from the plane $x-y+z=5$ measured along the line $x=y=z$ is :

3\sqrt{10}

10\sqrt3

\dfrac{10}{\sqrt3}

\dfrac{20}3

Answer and worked solution

Official answer: B

Along $(1+t,-5+t,9+t)$ , the plane gives $15+t=5$ , so $t=-10$ . The distance is $10\sqrt3$ .

JEE Main 2016 · 3 Apr · Shift 1, Mathematics question 83