The line of intersection of the planes $r . (3 i - j + k) = 1$ and $r . (i + 4 j - 2 k) = 2$ is parallel to the vector:

- 2 i + 7 j + 13 k

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2 i + 7 j - 13 k

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- 2 i - 7 j + 13 k

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2 i + 7 j + 13 k

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Solution

The correct option is B $- 2 i + 7 j + 13 k$
The line of intersection of the planes $r . (3 i - j + k) = 1$ and $r . (i + 4 j - 2 k) = 2$ is perpendicular to each of the normal vectors.
$n_{1} = 3 i - j + k$ and $n_{2} = i + 4 j - 2 k$
$∴$ It is parallel to the vector.
$n_{1} \times n_{2} = (3 i - j + k) \times (i + 4 j - 2 k) = - 2 i + 7 j + 13 k$

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Similar questions

\vec{r} \cdot (3 \hat{i} - \hat{j} + \hat{k}) = 1 and \vec{r} \cdot (\hat{i} + 4 \hat{j} - 2 \hat{k}) = 2

- 2 \hat{i} + 7 \hat{j} + 13 \hat{k}

2 \hat{i} + 7 \hat{j} - 13 \hat{k}

- 2 \hat{i} - 7 \hat{j} + 13 \hat{k}

2 \hat{i} + 7 \hat{j} + 13 \hat{k}

Check whether the two vectors, $\to A = - 3 ˆ i - 7 ˆ j + 9 ˆ k a n d \to B = - 2 ˆ i - 21 ˆ j + 6 ˆ k$ are parallel to each other

2 i - 7 j - k

4 j - 3 k

\to a = 2 ˆ i - 7 ˆ j - 3 ˆ k

A, B

C

- 2 i + j - k, - 4 i + 2 j + 2 k

6 i - 3 j - 13 k

A B = λ A C

λ

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