The equation of the plane passing through the point $(- 1, 2, 1)$ and perpendicular to the line joining the points $(- 3, 1, 2)$ and $(2, 3, 4)$ is ________.

¯ r \cdot (5^i + 2^j + 2^k) = 1

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¯ r \cdot (5^i + 2^j + 2^k) = - 1

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¯ r \cdot (5^i - 2^j + 2^k) = - 5

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¯ r \cdot (5^i - 2^j - 2^k) = 1

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Solution

The correct option is A $¯ r \cdot (5^i + 2^j + 2^k) = 1$
The parallel vector of line joining the points (-3,1,2) and (2,3,4) is $\to p$ = $5^i + 2^j + 2^k$

We want to find the plane perpendicular to $\to p$ , thus $\to p$ acts as normal vector to the plane .

We know that given a point on a plane with position vector (say $\to a$ ) and normal vector to plane ( say $\to n$ ), then Equation of plane is $(\to r - \to a) . (\to n) = 0$

So, Required equation of plane is { $\to r - (- \to i + 2 \to j + 1 \to k)} . (5 \to i + 2 \to j + 2 \to k) = 0 \to \to r . (5 \to i + 2 \to j + 2 \to k) = 1$

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Similar questions

¯ r - (^i - 2^j + 2^k) = 1

¯ r - (2^i -^j) + 2 = 0

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¯ r - (^i +^j + 2^k) + 9 = 0

L_{1} : ¯ r = 3^i -^j + 2^k + λ (2^i + 4^j -^k);

L_{2} : ¯ r =^i +^j - 3^k + μ (4^i + 2^j + 4^k);

L_{3} : ¯ r = 3^i - 2^j - 2^k + t (2^i + 4^j - 2^k)

(1, 6, 3)

¯ r = (^j + 2^k) + λ (^i + 2^j + 3^k)

¯ r = 2^i - 4^j + 2^k + λ (3^i + 4^j + 2^k)

¯ r \cdot (^i - 2^j +^k) = 0

\to r = 3^i + 2^j - 4^k + λ (^i + 2^j + 2^k)

\to r = 5^i - 2^j + μ (3^i + 2^j + 6^k)

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