Let + + = 1 be the equation of a plane passing through the point (3, - 2, 5) and perpendicular to the line joining the points (1, 2, 3) and (- 2, 3, 5). Then the value of α is equal to _____ .

Correct answer: 0 Given, + + = 1 be the equation of a plane passing through the point (3, - 2, 5) and perpendicular to the line joining the points (1, 2, 3) and (- 2, 3, 5), Now, normal vector of plane = (1 - (- 2) , 2 - 3, 3 - 5) ≡ (3, - 1, - 2) So, the equation of Plane will be, 3x - y - 2z + λ = 0 Now given, point (3, - 2, 5) satisfies the plane 3 × 3 + 2 - 10 + λ = 0 ⇒ λ = - 1 Hence, the equation of plane will be, 3x - y - 2z = 1 Now comparing with given equation of plane, we get α = 6

JEE Main 20231 FebruaryEvening Shift

Exam: JEE Main
Year: 2023
Date: 1 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2023 · 1 February · Evening Shift · Mathematics · Three Dimensional Geometry

Let + + = 1 be the equation of a plane passing through the point...

Let

\alphax + \betay + \gammaz = 1

be the equation of a plane passing through the point \left(3, - 2, 5\right)

and perpendicular to the line joining the points \left(1, 2, 3\right) and \left(- 2, 3, 5\right). Then the value of

\alpha\betay

is equal to _____ .

Correct answer

Step-by-step solution

Given,

$\alphax + \betay + \gammaz = 1$ be the equation of a plane passing through the point $\left(3, - 2, 5\right)$ and perpendicular to the line joining the points \left(1, 2, 3\right) and \left(- 2, 3, 5\right),

Now, normal vector of plane $= \left(1 - \left(- 2\right)$ , 2 - 3, 3 - 5\right) ≡ \left(3, - 1, - 2\right) $</p><p>So, the equation of Plane will be,$ 3x - y - 2z + \lambda = 0 $</p><p>Now given, point \left(3, - 2, 5\right)$ satisfies the plane

$3 \times 3 + 2 - 10 + \lambda = 0$

$⇒ \lambda = - 1$

Hence, the equation of plane will be, $3x - y - 2z = 1$

Now comparing with given equation of plane, we get $\alpha\betay = 6$

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