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Applications of Dot Product

Let a=a1i+a...

Question

Let $a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k$ and $c = c_{1} i + c_{2} j + c_{3} k$ be three non-zero vectors such that $c$ is a unit vector perpendicular to both $a$ and $b$ . If the angle between $a$ and $b$ is $π / 6$ , 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

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B

1

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C

\frac{1}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + c_{3}^{2})

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D

\frac{3}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + c_{3}^{2}) (c_{1}^{2} + c_{2}^{2} + c_{3}^{2})

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Solution

The correct option is C $\frac{1}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + c_{3}^{2})$
According to the given conditions,
$c_{1}^{2} + c_{2}^{2} + c_{3}^{2} = 1, a \cdot c = 0, b \cdot c = 0$
and $cos \frac{π}{6} = \frac{\sqrt{3}}{2} = \frac{a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}}{\sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}} \sqrt{b_{1}^{2} + b_{2}^{2} + b_{3}^{2}}}$
Thus $a_{1} c_{1} + a_{2} c_{2} + a_{3} c_{3} = 0, b_{1} c_{1} + b_{2} c_{2} + b_{3} c_{3} = 0$
and $\frac{\sqrt{3}}{2} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2})^{1 / 2} (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})^{1 / 2} = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$
Now,
${∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2} = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{2} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} a_{1}^{2} + a_{2}^{2} + a_{3}^{2} & a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} & a_{1} c_{1} + a_{2} c_{2} + a_{3} c_{3} a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} & b_{1}^{2} + b_{2}^{2} + b_{3}^{2} & b_{1} c_{1} + b_{2} c_{2} + b_{3} c_{3} a_{1} c_{1} + a_{2} c_{2} + a_{3} c_{3} & b_{1} c_{1} + b_{2} c_{2} + b_{3} c_{3} & c_{1}^{2} + c_{2}^{2} + c_{3}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} a_{1}^{2} + a_{2}^{2} + a_{3}^{2} & a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} & 0 a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} & b_{1}^{2} + b_{2}^{2} + b_{3}^{2} & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2}) - (a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3})^{2}$
$= (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})$
$- \frac{3}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})$
$= \frac{1}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})$

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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}

\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}

\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}

is equal to

(a) 0
(b) 1
(c)

\frac{1}{4} {|\vec{a}|}^{2} {|\vec{b}|}^{2}

\frac{3}{4} {|\vec{a}|}^{2} {|\vec{b}|}^{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} .

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, | c | = 1

(a \times b) \times c = 0

{∣ ∣ ∣ ∣ \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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