JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q84

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MathematicsSingle correct+4 / -1Question 84 of 90

The distance of the point (1, −2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes $x-y+2z=3$ and $2x-2y+z+12=0$ , is :

2\sqrt2

\sqrt2

\dfrac1{\sqrt2}

Answer and worked solution

Official answer: A

The normal is $(1,-1,2)\times(2,-2,1)=(3,3,0)\parallel(1,1,0)$ , so the plane is x + y = 3. The distance of (1, −2, 4) is $\frac{|1-2-3|}{\sqrt2}=2\sqrt2$ .

Topics:Equation of a plane Angles and distances

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JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 84