Let the points on the plane P be equidistant from the points (-4,2,1) and (2,-2,3). Then the acute angle between the plane P and the plane 2 x + y + 3 z =1 is

Question

Let the points on the plane P be equidistant from the points (-4,2,1) and (2,-2,3). Then the acute angle between the plane P and the plane 2 x + y + 3 z =1 is (A) (π/6) (B) (π/4) (C) (π/3) (D) (5 π/12)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/e1a3d231e88bbd981490f620675e2bc7-.png" /> Normal vector = AB =( OB - OA ) =(6 hat i -4 hat j +2 hat k ) or 2(3 hat i -2 hat j + hat k ) P ≡ 3( x +1)-2( y )+1( z -2)=0 P ≡ 3 x -2 y + z +1=0 P prime ≡ 2 x + y +3 z -1=0 angle between P P prime=|( hatn1 ⋅ hatn2/| n 1|| n 2|)|= cos θ θ= cos -1((6-2+3/√14 × √14)) θ= cos -1((7/14))= cos -1((1/2))=(π/3)

Q. Let the points on the plane $P$ be equidistant from the points $(- 4, 2, 1)$ and $(2, - 2, 3)$ . Then the acute angle between the plane $P$ and the plane $2 x + y +$ $3 z = 1$ is

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{5 π}{12}$

Q. Let the points on the plane P be equidistant from the points (−4,2,1) and (2,−2,3). Then the acute angle between the plane P and the plane 2x+y+ 3z=1 is

6π​

4π​

3π​

125π​

Normal vector =AB=(OB−OA) =(6i^−4j^​+2k^) or 2(3i^−2j^​+k^) P≡3(x+1)−2(y)+1(z−2)=0 P≡3x−2y+z+1=0 P′≡2x+y+3z−1=0 angle between P&P′=∣∣​∣n1​∣∣n2​∣n^1​⋅n^2​​∣∣​=cosθ θ=cos−1(14​×14​6−2+3​) θ=cos−1(147​)=cos−1(21​)=3π​

Q. Let the points on the plane $P$ be equidistant from the points $(- 4, 2, 1)$ and $(2, - 2, 3)$ . Then the acute angle between the plane $P$ and the plane $2 x + y +$ $3 z = 1$ is

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{5 π}{12}$