Consider the planes P1,P2 and points A(2, 5, 1) and B(1, 3, 0) where P1:x+y+z=3P2:2x+y−z=4Then which is/are CORRECT ?

(A) n¯1×n¯2=⟨−2,3,−1⟩Point common to P1 and P2 is (1, 2, 0)∴ distance between skew lines is(a¯−c¯)⋅(p¯×q¯)|p¯×q¯|=153BA→=i^+2j^+k^Angle between line and plane=sin−1⁡436=θProjection is |BA→|cos⁡θ=6×218=63 (D) r=263Area =r=26π3

Consider the planes P1,P2 and points A(2, 5, 1) and B(1, 3, 0) where P1:x+y+z=3P2:2x+y−z=4Then which is/are CORRECT ?

(A) n¯1×n¯2=⟨−2,3,−1⟩Point common to P1 and P2 is (1, 2, 0)∴ distance between skew lines is(a¯−c¯)⋅(p¯×q¯)|p¯×q¯|=153BA→=i^+2j^+k^Angle between line and plane=sin−1⁡436=θProjection is |BA→|cos⁡θ=6×218=63 (D) r=263Area =r=26π3

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Consider the planes $P_{1}, P_{2}$ and points A(2, 5, 1) and B(1, 3, 0) where
$\begin{aligned} P_{1} : x + y + z = 3 \\ P_{2} : 2 x + y - z = 4 \end{aligned}$
Then which is/are CORRECT ?

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Shortest distance between line AB and points common to $P_{1}, P_{2}$ is equal to $\frac{1}{5 \sqrt{3}}$ units.

Projection of segment AB on P₁ is equal to $\frac{\sqrt{6}}{3}$ units

Projection of segment AB on P₁ is equal to $2 \sqrt{2} units$

Area enclosed by locus of all points on P₁ which are at a distance of 3 units from B is equal to $\frac{26 π}{3} sq$ . units.

answer is A, B, D.

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Detailed Solution

$(A) {\bar{n}}_{1} \times {\bar{n}}_{2} = ⟨ - 2, 3, - 1 ⟩$
Point common to P1 and P2 is (1, 2, 0)
$∴$ distance between skew lines is
$|\frac{(\bar{a} - \bar{c}) \cdot (\bar{p} \times \bar{q})}{| \bar{p} \times \bar{q} |}| = \frac{1}{5 \sqrt{3}}$
Question Image
$\vec{BA} = \hat{i} + 2 \hat{j} + \hat{k}$
Angle between line and plane
$= \sin^{- 1} (\frac{4}{\sqrt{3} \sqrt{6}}) = θ$
Projection is $| \vec{BA} | \cos θ = \frac{\sqrt{6} \times \sqrt{2}}{\sqrt{18}} = \frac{\sqrt{6}}{3}$
$(D) r = \sqrt{\frac{26}{3}}$
Area $= r = \frac{26 π}{3}$ Question Image