If bar{a}=a_1 hat{i}+a_2 hat{j}+a_3 hat{k}, b | vedclass

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(B) First,note that the determinant on the right side is equal to the scalar triple product $[\bar{a} \quad \bar{b} \quad \bar{c}]$.$\begin{vmatrix} a

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(B) First,note that the determinant on the right side is equal to the scalar triple product $[\bar{a} \quad \bar{b} \quad \bar{c}]$.$\begin{vmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{vmatrix} = [\bar{a} \quad \bar{b} \quad \bar{c}]$.Now,calculate the scalar triple product on the left side:$[3 \bar{a}+\bar{b} \quad 3 \bar{b}+\bar{c} \quad 3 \bar{c}+\bar{a}] = (3 \bar{a}+\bar{b}) \cdot ((3 \bar{b}+\bar{c}) 	imes (3 \bar{c}+\bar{a}))$.Expanding the cross product:$(3 \bar{b}+\bar{c}) 	imes (3 \bar{c}+\bar{a}) = 9(\bar{b} 	imes \bar{c}) + 3(\bar{b} 	imes \bar{a}) + 3(\bar{c} 	imes \bar{c}) + (\bar{c} 	imes \bar{a}) = 9(\bar{b} 	imes \bar{c}) + 3(\bar{b} 	imes \bar{a}) + (\bar{c} 	imes \bar{a})$.Now take the dot product with $(3 \bar{a}+\bar{b})$:$= (3 \bar{a}+\bar{b}) \cdot (9(\bar{b} 	imes \bar{c}) + 3(\bar{b} 	imes \bar{a}) + (\bar{c} 	imes \bar{a}))$$= 27[\bar{a} \quad \bar{b} \quad \bar{c}] + 9[\bar{a} \quad \bar{b} \quad \bar{a}] + 3[\bar{a} \quad \bar{c} \quad \bar{a}] + 9[\bar{b} \quad \bar{b} \quad \bar{c}] + 3[\bar{b} \quad \bar{b} \quad \bar{a}] + [\bar{b} \quad \bar{c} \quad \bar{a}]$.Since any scalar triple product with two identical vectors is $0$,this simplifies to:$= 27[\bar{a} \quad \bar{b} \quad \bar{c}] + [\bar{b} \quad \bar{c} \quad \bar{a}] = 27[\bar{a} \quad \bar{b} \quad \bar{c}] + [\bar{a} \quad \bar{b} \quad \bar{c}] = 28[\bar{a} \quad \bar{b} \quad \bar{c}]$.Comparing this with $\lambda [\bar{a} \quad \bar{b} \quad \bar{c}]$,we get $\lambda = 28$.

Similar Questions

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