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Bisector of Angle between Two Vectors

Let a →=a 1 i...

Question

Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} and \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ be three non-zero vectors such that $\vec{c}$ is a unit vector perpendicular to both $\vec{a} and \vec{b}$ . If the angle between $\vec{a} and \vec{b}$ is $\frac{π}{6},$ then ${|\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}|}^{2}$ is equal to

(a) 0
(b) 1
(c) $\frac{1}{4} {|\vec{a}|}^{2} {|\vec{b}|}^{2}$

(d) $\frac{3}{4} {|\vec{a}|}^{2} {|\vec{b}|}^{2}$

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Solution

(c) $\frac{1}{4} {|\vec{a}|}^{2} {|\vec{b}|}^{2}$

$We have {|\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}|}^{2} = {[(\vec{a} \times \vec{b}) . \vec{c}]}^{2} (By definition of scalar triple product) = {[|(\vec{a} \times \vec{b})| |\vec{c}| c os 0 °]}^{2} (∵ \vec{a} \times \vec{b} is parallel to vector \vec{c} as \vec{c} is perpendicular to both \vec{a} and \vec{b}) = {(|\vec{a}| |\vec{b}| s in \frac{π}{6})}^{2} (∵ |\vec{c}| = 1 and angle between \vec{a} and \vec{b} is \frac{π}{6}) = {|\vec{a}|}^{2} {|\vec{b}|}^{2} {(\frac{1}{2})}^{2} = \frac{1}{4} {|\vec{a}|}^{2} {|\vec{b}|}^{2}$

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Similar questions

a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k

c = c_{1} i + c_{2} j + c_{3} k

be three non-zero such that

c

is a unit perpendicular to both vectors

a

b

. If the angle between vectors

a

b

\frac{π}{6},

{∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2}

a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k,

c = c_{1} i + c_{2} j + c_{3} k

be three non-zero vectors such that

c

is unit vector perpendicular to both vectors

a

b

. If the angle between vectors

a

b

\frac{π}{6}

{∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2}

a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k

c = c_{1} i + c_{2} j + c_{3} k

be three non-zero vectors such that

c

is a unit vector perpendicular to both

a

b

. If the angle between

a

b

π / 6

{∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2}

\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} and \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k},

then verify that

\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} .

Q. If three non-zero vectors are

a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k

c = c_{1} i + c_{2} j + c_{3} k

If c is the unit vector perpendicular to the vectors a and b and the angle between a and b is

\frac{π}{6}

{∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2}

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