Let the vectors a- b- c- given as a1-a2-a3k- b1-b2-b3k- c1-c2-c3k- Then show that -a- b-c-a-b-a-c- -

We have, a⃗=a_1î+a_2ĵ+a_3k̂, #10;b=b_1î+b_2k̂+b_3k̂, c⃗=c_1#10;i+c_2ĵ+c_3k̂ (b⃗+c⃗)=(b_1+c_1)î+(b_2+c_2)ĵ+(b_3+c_3)k̂ Now, #10; a⃗× (b⃗+c⃗) î ...

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Byju's Answer

Standard XII

Mathematics

Theorems on Integration

Let the vecto...

Question

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

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Solution

We have,
$\to a = a_{1}^i + a_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^k + b_{3}^k, \to c = c_{1}^i + c_{2}^j + c_{3}^k$
$(\to b + \to c) = (b_{1} + c_{1})^i + (b_{2} + c_{2})^j + (b_{3} + c_{3})^k$
Now, $\to a \times (\to b + \to c) ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k a_{1} & a_{2} & a_{3} b_{1} + c_{1} & b_{2} + c_{2} & b_{3} + c_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$=^i [a_{2} (b_{3} + c_{3}) - a_{3} (b_{2} + c_{2})] -^j [a_{1} (b_{3} + c_{3} - a_{3} (b_{1} + c_{1})] +^k [a_{1} (b_{2} + c_{2} - a_{2} (b_{1} + c_{1})]$
$=^i [a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}] +^j [- a_{1} b_{3} - a_{1} c_{3} + a_{3} b_{1} + a_{3} c_{1}] +^k [a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{1}] . . . . . . . . . . . . . (1)$
$\to a \times \to b = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$=^i [a_{2} b_{3} - a_{3} b_{2}] +^j [b_{1} a_{3} - a_{1} b_{3}] +^k [a_{1} b_{2} - a_{2} b_{1}] . . . . . . . . . . (2)$
$\to a \times \to c = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k a_{1} & a_{2} & a_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$=^i [a_{2} c_{3} - a_{3} c_{2}] +^j [a_{3} c_{1} - a_{1} c_{3}] +^k [a_{1} c_{2} - a_{2} c_{1}] . . . . . . . . . . . (3)$
On adding $(2)$ and $(3)$ , we get:
$(\to a \times \to b) + (\to a \times \to c)$
$=^i [a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}] +^j [b_{1} a_{3} + a_{3} c_{1} - a_{1} b_{3} - a_{1} c_{3}] +^k [a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{1} - a_{2} c_{1}] . . . . . . . . . . (4)$
Now, from $(1)$ and $(4)$ , we have:
$\to a \times (\to b + \to c) = \to a \times \to b + \to a \times \to c$
Hence, the given result is proved.

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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 $\frac{π}{6},$
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