Value of - such that the line x-12-y-13-z-1- is - to normal to the plane r- 2i-3j-4k-0 is-

The correct option is A 134 Since, the line x 1 2 = y 1 3 = z 1 λ #160; is perpendicular to normal of the plane r.(2i+3j+4k)=0 Therefore ...

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Byju's Answer

Standard XII

Mathematics

Angle between Two Planes

Value of λ ...

Question

Value of $λ$ such that the line $\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{λ}$ is $⊥$ to normal to the plane $\to r \cdot (2 \to i + 3 \to j + 4 \to k) = 0$ is:

- \frac{13}{4}

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- \frac{17}{4}

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None of these.

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Solution

The correct option is A $- \frac{13}{4}$
Since, the line $\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{λ}$ is perpendicular to normal of the plane $r . (2 i + 3 j + 4 k) = 0$
Therefore, $(2 i + 3 j + λ k) . (2 i + 3 j + 4 k) = 0$
$\Rightarrow 4 + 9 + 4 λ = 0 \Rightarrow λ = \frac{- 13}{4}$

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lf the angle $θ$ between the line $\frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ and the plane $2 x - y + \sqrt{λ} z + 4 =$ $0$ is such that $sin θ = \frac{1}{3}$ , then value of $λ$ is

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Value of λ such that the line x−12=y−13=z−1λ is ⊥ to normal to the plane →r⋅(2→i+3→j+4→k)=0 is:

The correct option is A −134Since, the line x−12=y−13=z−1λ is perpendicular to normal of the plane r.(2i+3j+4k)=0Therefore,(2i+3j+λk).(2i+3j+4k)=0 ⇒4+9+4λ=0⇒λ=−134

Value of $λ$ such that the line $\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{λ}$ is $⊥$ to normal to the plane $\to r \cdot (2 \to i + 3 \to j + 4 \to k) = 0$ is:

The correct option is A $- \frac{13}{4}$
Since, the line $\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{λ}$ is perpendicular to normal of the plane $r . (2 i + 3 j + 4 k) = 0$
Therefore, $(2 i + 3 j + λ k) . (2 i + 3 j + 4 k) = 0$
$\Rightarrow 4 + 9 + 4 λ = 0 \Rightarrow λ = \frac{- 13}{4}$