JEE Main 2022 · 30 Jul · Shift 2 · Mathematics Q77

← JEE Main 2022 · 30 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 77 of 90

Let $ax+by+cz+3=0$ be the plane which bisects the acute angle between the planes $2x-y-2z+3=0$ and $3x-2y+6z+8=0$ . Then a + b + c is equal to

−31

\dfrac{14}{15}

−28

Answer and worked solution

Official answer: B

Bisectors $\dfrac{2x-y-2z+3}3=\pm\dfrac{3x-2y+6z+8}7$ : $23x-13y+4z+45=0$ or $5x-y-32z-3=0$ .

Acute one. The planes meet at $\cos^{-1}\frac4{21}\approx79°$ , so the acute bisector makes about 39.5° with each plane. $(5,-1,-32)$ makes $\cos^{-1}\frac{75}{3\sqrt{1050}}\approx39.5°$ with $(2,-1,-2)$ ; the other makes about 50.5°.

Scale to constant +3: $-5x+y+32z+3=0$ , so $a+b+c=28$ .

(B)

Topics:Equation of a plane Angles and distances

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JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 77