If 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 and a - b - c - 0 then - - a 1 a 2 a 3- b 1 b 2 b 3- c 1 c 2 c 3 - 2 is equal to

The correct option is B 0Given that ( a× amp; b)× #160; amp; c =0 ( a amp; b amp; c ) #160; ^ 2 =( a amp; b amp; c ) ×( a ...

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Byju's Answer

Standard XII

Mathematics

Adjoint of a Matrix

If a = a 1...

Question

If $a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k c = c_{1} i + c_{2} j + c_{3} k, | c | = 1$ and $(a \times b) \times c = 0$ 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

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| a |^{2} | b |^{2}

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| a \times b |^{2}

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Solution

The correct option is B 0
Given that $(\begin{matrix} a \times & b) \times & c \end{matrix} = 0$

${(\begin{matrix} a & b & c \end{matrix})}^{2} = (\begin{matrix} a & b & c \end{matrix}) \times (\begin{matrix} a & b & c \end{matrix}) - - - - (1)$

Using cross product of two vectors,
$(\begin{matrix} u_{1} & u_{2} & u_{3} \end{matrix}) \times (\begin{matrix} v_{1} & v_{2} & v_{3} \end{matrix}) = \times (\begin{matrix} u_{2} v_{3} - u_{3} v_{2} & u_{3} v_{1} - u_{1} v_{3} & u_{1} v_{2} - u_{2} v_{1} \end{matrix})$

Now applying this in the equation (1), we get

${(\begin{matrix} a & b & c \end{matrix})}^{2} = (\begin{matrix} a & b & c \end{matrix}) \times (\begin{matrix} a & b & c \end{matrix}) = (\begin{matrix} 0 & 0 & 0 \end{matrix}) (g i v e n)$

Hence option (a) is correct.

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Similar questions

Q. If

\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. 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 such that

c

is a unit perpendicular to both vectors

a

and

b

. If the angle between vectors

a

and

b

\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

Q. 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

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

Q. 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}

\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}

Q. 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 unit vector perpendicular to both vectors

a

and

b

. If the angle between vectors

a

and

b

\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

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If a=a1i+a2j+a3k,b=b1i+b2j+b3kc=c1i+c2j+c3k,|c|=1 and (a×b)×c=0 then ∣∣ ∣∣a1a2a3b1b2b3c1c2c3∣∣ ∣∣2 is equal to

The correct option is B 0Given that (a×b)×c=0(abc)2=(abc)×(abc)−−−−(1)Using cross product of two vectors,(u1u2u3)×(v1v2v3)=×(u2v3−u3v2u3v1−u1v3u1v2−u2v1) Now applying this in the equation (1), we get(abc)2=(abc)×(abc)=(000)(given)Hence option (a) is correct.

If $a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k c = c_{1} i + c_{2} j + c_{3} k, | c | = 1$ and $(a \times b) \times c = 0$ 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