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Point Form of Normal:Hyperbola

If a vector '...

Question

If a vector 'r' of magnitude $3 \sqrt{6}$ is directed along the bisector of the angle between the vectors a = 7i - 4j - 4 k and b = -2i -j + 2k, then r is equal to

A

i - 7j + 2k

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B

i + 7j - 2k

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C

i + 7j + 2k

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D

i - 7j - 2k

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Solution

The correct option is A i - 7j + 2k
$\frac{7 x - 4 y - 4 k}{\sqrt{81}} = 1 [\frac{- 2 x - y + 2 z}{\sqrt{3}}]$
$\frac{7 x - 4 y - 4 k}{9} = 1 [\frac{- 2 x - 4 + 2 z}{3}]$
$7 x - 4 y - 4 k = \pm [- 6 x - 3 y + 6 z]$
takes the
$7 x - 4 y - 4 k = - 6 x - 3 y + 6 z$
$13 x - y - 10 k = 0$
take $-^{'} v e^{'}$
$7 x - 4 y - 4 k = 6 x + 3 y - 6 z$
$x - 7 y + 2 z = 0$
magnitude $= \sqrt{1 + 49 + 4} = \sqrt{54} = 2 \sqrt{13}$
So it should be it is Answer

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Similar questions

Q. A vector parallel to the line of intersection of the planes

\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}

r

3 \sqrt{6}

is directed along the bisector of the angle between the vectors

a = 7 i - 4 j - 4 k

b = - 2 i - j + 2 k

r

\to r

3 \sqrt{6}

is directed along the bisector of the angle between the vectors

\to a = 7 ˆ i - 4 ˆ j - 4 ˆ k

\to b = - 2 ˆ i - ˆ j + 2 ˆ k

\to r =

\to c

5 \sqrt{6}

directed along the bisector of the angle between

\to a = 7 ˆ i - 4 ˆ j - 4 ˆ k

\to b = - 2 ˆ i - ˆ j + 2 ˆ k

c

, directed along the internal bisector of the angle between the vectors

a = 7 i - 4 j - 4 k

b = - 2 i - j + 2 k,

| c | = 5. \sqrt{6},

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