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Condition for Two Lines to Be Perpendicular

From a point ...

Question

From a point $P (λ, λ, λ)$ , perpendicular $P Q$ and $P R$ are drawn respectively on the lines $y = x, z = 1$ and $y = - x, z = - 1$ . If $P$ is such that $∠ Q P R$ is a right angle, then the possible value(s) of $λ$ is(are)

\sqrt{2}

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- \sqrt{2}

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Solution

The correct option is C $- 1$

DR's of $P Q = (0, 0, λ - 1) = (λ - 1)^k$

$P R = (λ, λ, λ + 1) = (λ^i + λ^j + (λ + 1)^k)$

$P Q ⊥ P R = - - \to P Q . - - \to P R = λ^{2} - 1 = 0$

$λ = \pm 1$

$λ = 1$ gives same value of $P$ and $Q$

$∴ λ = - 1$

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Similar questions

From a point $P (λ, λ, λ)$ , perpendiculars PQ and PR are drawn respectively on the lines y = x, z = 1 and y = -x, z = -1. If P is such that $∠ Q P R$ is a right angle, then the possible value(s) of
$λ$ is (are)

Q. From a point

P (λ, λ, λ)

, perpendiculars PQ and PR are drawn perpendiculars on the lines

y = x, z = 1 . a n d y = - x, z = - 1

. If P is such that

∠ Q P R

is a right angle, then the possible values(s) of

λ

is (are) :

Q. From a point

P (λ, λ, λ)

, perpendiculars

P Q

and

P R

are drawn respectively on the lines

y = x, z = 1

and

y = - x, z = - 1

. If

P

is such that

∠ Q P R

is a right angle, then the possible value(s) of

λ

is(are)

Q. If the lines

x = 1 + a, y = - 3 - λ a, z = 1 + λ a

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are coplanar, then

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is equal to

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From a point P(λ,λ,λ), perpendicular PQ and PR are drawn respectively on the lines y=x,z=1 and y=−x,z=−1. If P is such that ∠QPR is a right angle, then the possible value(s) of λ is(are)

The correct option is C −1DR's of PQ=(0,0,λ−1)=(λ−1)^kPR=(λ,λ,λ+1)=(λ^i+λ^j+(λ+1)^k)PQ⊥PR=−−→PQ.−−→PR=λ2−1=0λ=±1λ=1 gives same value of P and Q∴λ=−1

From a point $P (λ, λ, λ)$ , perpendicular $P Q$ and $P R$ are drawn respectively on the lines $y = x, z = 1$ and $y = - x, z = - 1$ . If $P$ is such that $∠ Q P R$ is a right angle, then the possible value(s) of $λ$ is(are)

The correct option is C $- 1$

DR's of $P Q = (0, 0, λ - 1) = (λ - 1)^k$

$P R = (λ, λ, λ + 1) = (λ^i + λ^j + (λ + 1)^k)$

$P Q ⊥ P R = - - \to P Q . - - \to P R = λ^{2} - 1 = 0$

$λ = \pm 1$

$λ = 1$ gives same value of $P$ and $Q$

$∴ λ = - 1$