If a-3k-b-k- c-2- such that a-b-1 and b-c-3then 1-3a-b- c is equal to

Finding the value of the expression:Step : Finding the value of a×b:Given,a=(α , β ,3), 0ex0exb=( β , α , 1),0ex0exc = (1, 2, 1),a·b=10ex0ex⇒ ( αβ αβ 3)=10 ...

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Byju's Answer

Standard VII

Mathematics

Operations on Rational Numbers using PEMDAS Rule

If a=αî+βĵ+3k...

Question

If $\vec{a} = α \hat{i} + β \hat{j} + 3 k, \overset{⇀}{b} = - β \hat{i} - α \hat{j} - k$ , $\vec{c} = \hat{i} - 2 \hat{j}$ such that $\vec{a} . \vec{b} = 1$ and $\vec{b .} \vec{c} = - 3$ then $\frac{1}{3} ((\vec{a} \times \vec{b}) . \vec{c})$ is equal to

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Solution

Finding the value of the expression:
Step : Finding the value of $\vec{a} \times \vec{b}$ :
Given,
$\vec{a} = (α, β, 3), \vec{b} = (- β, - α, - 1), \vec{c} = (1, - 2, - 1),$
$\vec{a} \cdot \vec{b} = 1 \Rightarrow (- α β - α β - 3) = 1 \Rightarrow - 2 α β = 4 \Rightarrow α β = - 2 . . . (1)$
$\vec{b} . \vec{c} = - 3 \Rightarrow (- β + 2 α + 1) = - 3 \Rightarrow 2 α - β = - 4 . . . (2)$
Solving the equations $(1)$ and $(2)$ , we get
$α = - 1 β = 2$
So, vector $\vec{a} = (- 1, 2, 3)$ and $\vec{b} = (- 2, 1, - 1)$
$\vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 2 & 3 \\ - 2 & 1 & - 1 \end{matrix}| = \hat{i} (- 5) - \hat{j} (7) + k (3) = (- 5, - 7, 3)$
Furthermore,
$(\vec{a} \times \vec{b}) . \vec{c} = (- 5, - 7, 3) . (1, - 2, - 1) = - 5 + 14 - 3 = 6$
$\frac{1}{3} ((\vec{a} x \vec{b}) . \vec{c}) = 2$
Hence, the required answer is 2.

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If a→=αi^+βj^+3k,b⇀=–βi^–αj^–k, c→=i^-2j^ such that a→.b→=1 and b.→c→=-3then 13((a→×b→).c→) is equal to

If $\vec{a} = α \hat{i} + β \hat{j} + 3 k, \overset{⇀}{b} = - β \hat{i} - α \hat{j} - k$ , $\vec{c} = \hat{i} - 2 \hat{j}$ such that $\vec{a} . \vec{b} = 1$ and $\vec{b .} \vec{c} = - 3$ then $\frac{1}{3} ((\vec{a} \times \vec{b}) . \vec{c})$ is equal to