Let S be the reflection of a point Q with respect to the plane given by → r = - (t + p) ^ i + t ^ j + (1 + p) ^ k where t, p are real parameters and ^ i , ^ j , ^ k are the unit vectors along the three positive

Correct answer: A. 3(α + β) = - 101 Given, Equation of plane is → r = - (t + p) ^ i + t ^ j + (1 + p) ^ k On rearranging we get, → r = ^ k + t(- ^ i + ^ j ) + p(- ^ i + ^ k ) So, equation of plane in standard form is given by [ cases → r - ^ k & - ^ i + ^ j & - ^ i + ^ k cases ] = 0 ⇒ | cases x & y & z - 1 - 1 & 1 & 0 - 1 & 0 & 1 cases | = 0 ⇒ x + y + z = 1 ⋯ (1) Now given coordinate of Q = (10, 15, 20) And coordinates of S = (α, β, γ) Now using the image formula of point and plane we get, (α - 10)/(1) = (β - 15)/(1) = (γ - 20)/(1) = (- 2(10 + 15 + 20 - 1))/(3) ⇒ α - 10 = β - 15 = γ - 20 = - (88)/(3) ⇒ α = - (58)/(3), β = - (43)/(3), γ = - (28)/(3) Now solving all options we get, 3(α + β) = - 101 3(β + γ) = - 71 3(γ + α) = - 86 3(α + β + γ) = - 129

JEE Advanced 2022Paper 1

Exam: JEE Advanced
Year: 2022
Paper: Paper 1
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Advanced 2022 · Paper 1 · Mathematics · Three Dimensional Geometry

Let S be the reflection of a point Q with respect to the plane...

Let

S

be the reflection of a point

Q

with respect to the plane given by

\overset{\to}{r} = - \left(t + p\right)\overset{^}{i} + t\overset{^}{j} + \left(1 + p\right)\overset{^}{k}

where

t, p

are real parameters and

\overset{^}{i}, \overset{^}{j}, \overset{^}{k}

are the unit vectors along the three positive coordinate axes. If the position vectors of

Q

and

S

are

10\overset{^}{i} + 15\overset{^}{j} + 20\overset{^}{k}

and

\alpha\overset{^}{i} + \beta\overset{^}{j} + \gamma\overset{^}{k}

respectively, then which of the following is/are TRUE?

Options

A $3\left(\alpha + \beta\right) = - 101$
B $3\left(\beta + \gamma\right) = - 71$
C $3\left(\gamma + \alpha\right) = - 86$
D $3\left(\alpha + \beta + \gamma\right) = - 121$

Correct answer

A. $3\left(\alpha + \beta\right) = - 101$

Step-by-step solution

Given,

Equation of plane is

$\overset{\to}{r} = - \left(t + p\right)\overset{^}{i} + t\overset{^}{j} + \left(1 + p\right)\overset{^}{k}$

On rearranging we get,

$\overset{\to}{r} = \overset{^}{k} + t\left(- \overset{^}{i} + \overset{^}{j}\right) + p\left(- \overset{^}{i} + \overset{^}{k}\right)$

So, equation of plane in standard form is given by

\left[ $\begin{cases}\overset{\to}{r} - \overset{^}{k} & - \overset{^}{i} + \overset{^}{j} & - \overset{^}{i} + \overset{^}{k}\end{cases}$ \right] = 0 $$ ⇒ \left| $\begin{cases}x & y & z - 1 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1\end{cases}$ \right| = 0 $$ ⇒ x + y + z = 1 ⋯ \left(1\right) $Now given coordinate of$ Q = \left(10, 15, 20\right)

And coordinates of $S = \left(\alpha, \beta, \gamma\right)$

Now using the image formula of point and plane we get,

$\dfrac{\alpha - 10}{1} = \dfrac{\beta - 15}{1} = \dfrac{\gamma - 20}{1} = \dfrac{- 2\left(10 + 15 + 20 - 1\right)}{3}$

$⇒ \alpha - 10 = \beta - 15 = \gamma - 20 = - \dfrac{88}{3}$

$⇒ \alpha = - \dfrac{58}{3}, \beta = - \dfrac{43}{3}, \gamma = - \dfrac{28}{3}$

Now solving all options we get,

$3\left(\alpha + \beta\right) = - 101$

$3\left(\beta + \gamma\right) = - 71$

$3\left(\gamma + \alpha\right) = - 86$

$3\left(\alpha + \beta + \gamma\right) = - 129$

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