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Byju's Answer

Standard XII

Mathematics

Summation of Determinant

If a →=a 1 i∧...

Question

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

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Solution

$Given : \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} \vec{b} + \vec{c} = (b_{1} + c_{1}) \hat{i} + (b_{2} + c_{2}) \hat{j} + (b_{3} + c_{3}) \hat{k} ∴ \vec{a} \times (\vec{b} + c) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} + c_{1} & b_{2} + c_{2} & b_{3} + c_{3} \end{matrix}| = (a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}) \hat{i} - (a_{1} b_{3} + a_{1} c_{3} - a_{3} b_{1} - a_{3} c_{1}) \hat{j} + (a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{1}) \hat{k} . . . (1) Now, \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix}| = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} - (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \vec{a} \times \vec{c} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}| = (a_{2} c_{3} - a_{3} c_{2}) \hat{i} - (a_{1} c_{3} - a_{3} c_{1}) \hat{j} + (a_{1} c_{2} - a_{2} c_{1}) \hat{k} \vec{a} \times \vec{b} + \vec{b} \times \vec{c} = (a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}) \hat{i} - (a_{1} b_{3} + a_{1} c_{3} - a_{3} b_{1} - a_{3} c_{1}) \hat{j} + (a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{1}) \hat{k} . . . (2) From (1) and (2), we get \vec{a} \times (\vec{b} + c) = \vec{a} \times \vec{b} + \vec{b} \times \vec{c}$

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Similar questions

Q. 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

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

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

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}

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If a →=a1i^+a2j^+a3k^, b →=b1i^+b2j^+b3k^ and c →=c1i^+c2j^+c3k^, then verify that a →× b →+c →=a →×b →+a →×c →.

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