If a- - a 1 - - a 2 - - a 3 k- and b- - b 1 - - b 2 - - b 3 k- then find the value of a-b-

Given: a⃗ = a _ 1 î + a _ 2 ĵ a _ 3 k̂ and b⃗ = b _ 1 î + b _ 2 ĵ b _ 3 k̂ a⃗×b⃗ =î #160; amp; ĵ #160; amp; k̂ #160; a _ 1 ...

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Byju's Answer

Standard XII

Mathematics

Finding Inverse Using Elementary Transformations

If a⃗ = a 1...

Question

If $\to a = a_{1}^i + a_{2}^j - a_{3}^k$ and $\to b = b_{1}^i + b_{2}^j - b_{3}^k$ then find the value of $\to a \times \to b$ .

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Solution

Given: $\to a = a_{1}^i + a_{2}^j - a_{3}^k$ and $\to b = b_{1}^i + b_{2}^j - b_{3}^k$
$\to a \times \to b = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ =^i ∣ ∣ ∣ \begin{matrix} a_{2} & a_{3} b_{2} & b_{3} \end{matrix} ∣ ∣ ∣ -^j ∣ ∣ ∣ \begin{matrix} a_{1} & a_{3} b_{1} & b_{3} \end{matrix} ∣ ∣ ∣ +^k ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} b_{1} & b_{2} \end{matrix} ∣ ∣ ∣$
$= (a_{2} b_{3} - a_{3} b_{2})^i + (a_{1} b_{3} - a_{3} b_{1})^j - (a_{1} b_{2} - a_{2} b_{1})^k$

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Similar questions

Q. Let the vectors

\to a, \to b, \to c

given as

a_{1}^i + a_{2}^j + a_{3}^k, b_{1}^i + b_{2}^j + b_{3}^k, c_{1}^i + c_{2}^j + c_{3}^k

. Then show that

= \to a \times (\to b + \to c) = \to a \times \to b + \to a \times \to c

Q. The vector

a_{1}^i + a_{2}^j + a_{3}^k

and

b_{1}^i + b_{2}^j + b_{3}^k

are perpendicular to each other if :

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

Q. If

\to a = a_{1}^i + a_{2}^j + a_{3}^k; \to b = b_{1}^i + b_{2}^j + b_{3}^k;

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

and

[3 \to a + \to b, 3 \to b + \to c, 3 \to c + \to a]

= λ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to a \cdot^i & \to a \cdot^j & \to a \cdot^k \to b \cdot^i & \to b \cdot^j & \to b \cdot^k \to c \cdot^i & \to c \cdot^j & \to c \cdot^k \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

then the value of

\frac{λ}{4}

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If →a=a1^i+a2^j−a3^k and →b=b1^i+b2^j−b3^k then find the value of →a×→b.

Given: →a=a1^i+a2^j−a3^k and →b=b1^i+b2^j−b3^k→a×→b=∣∣ ∣ ∣∣^i^j^ka1a2a3b1b2b3∣∣ ∣ ∣∣=^i∣∣∣a2a3b2b3∣∣∣−^j∣∣∣a1a3b1b3∣∣∣+^k∣∣∣a1a2b1b2∣∣∣=(a2b3−a3b2)^i+(a1b3−a3b1)^j−(a1b2−a2b1)^k

If $\to a = a_{1}^i + a_{2}^j - a_{3}^k$ and $\to b = b_{1}^i + b_{2}^j - b_{3}^k$ then find the value of $\to a \times \to b$ .