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Bisector of Angle between Two Vectors

A vector c⃗...

Question

A vector $\to c$ of magnitude $5 \sqrt{6}$ directed along the bisector of the angle between $\to a = 7 ˆ i - 4 ˆ j - 4 ˆ k$ and $\to b = - 2 ˆ i - ˆ j + 2 ˆ k$ , is

\pm \frac{5}{3} (2 ˆ i + 7 ˆ j + ˆ k)

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\pm \frac{3}{5} (ˆ i + 7 ˆ j + 2 ˆ k)

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\pm \frac{5}{3} (ˆ i - 2 ˆ j + 7 ˆ k)

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\pm \frac{5}{3} (ˆ i - 7 ˆ j + 2 ˆ k)

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Solution

The correct option is C $\pm \frac{5}{3} (ˆ i - 7 ˆ j + 2 ˆ k)$
The angle bisectors will be sum and difference of the unit vectors of the two arms.

Thus, $\frac{7^i - 4^j - 4^k}{9} \pm \frac{- 2^i -^j + 2^k}{3}$

$= \frac{^i - 7^j + 2^k}{9}$ or $\frac{15^i -^j - 10^k}{9}$

We find the first direction in the options and since we need a vector of magnitude $5 \sqrt{6}$ , we have $\pm \frac{5}{3} (^i - 7^j + 2^k)$

Suggest Corrections

Similar questions

\to c = - 5 ˆ i + 7 ˆ j + 2 ˆ k

\to a = ˆ i + 2 ˆ j + 2 ˆ k

\to b = - 8 ˆ i + ˆ j - 4 ˆ k .

\to c

\to a

\to b

\to r

3 \sqrt{6}

\to a = 7 ˆ i - 4 ˆ j - 4 ˆ k

\to b = - 2 ˆ i - ˆ j + 2 ˆ k

\to r =

\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}

The vector c directed along the internal bisector of the angle between the vectors a = 7i - 4j - 4k and b = -2i - j + 2k with |c| = 5√6 ?

c

a = 7 i - 4 j - 4 k

b = - 2 i - j + 2 k,

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

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