If a=i+j, b=j+k, c=i+k , then ([(b × c) ×(c × a)(c × a) ×(a × b)(a × b) ×(b × c)])/([b+c c+a a+b][b × c × a a × b])

Correct answer: (B) 1. Since, (b × c) ×(c × a)=[a b c] c (c × a) ×(a × b)=[a b c] a (a × b) ×(b × c)=[a b c] b and [b+c c+a a+b]=2[a · b · c] and [b × c ~ c × a a × b]=[a b c]^2 [(b × c) ×(c × a)(c × a) ×(a × b) So, ((a × b) ×(b × c)])/([b+c c+a a+b][b × c × a a × b]) =([[a b c] c[a b c] a[a b c] b])/(2[a b c][a b]^2) =([a b c]^3[a b c])/(2[a b c]^3)=([a b c])/2=1 ⌈∵[a b c]= | 1 1 0 0 1 1 1 0 1 |=2 ⌋

Q74 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

If $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{k}}$, then $\frac{[(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})(\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})]}{[\mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a} \mathbf{a}+\mathbf{b}][\mathbf{b} \times \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]}$

A
0
Correct answer
B
1
C
$\sqrt{3}$
D
$\sqrt{2}$

Solution

Since, $(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{c}$ $$ \begin{aligned} & (\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{a} \\ & (\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{b} \\ & \text { and }[\mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a} \mathbf{a}+\mathbf{b}]=2[\mathbf{a} \cdot \mathbf{b} \cdot \mathbf{c}] \\ & \text { and }[\mathbf{b} \times \mathbf{c ~ c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]=[\mathbf{a} \mathbf{b} \mathbf{c}]^2 \\ & {[(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})} \\ & \text { So, } \frac{(\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})]}{[\mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a} \mathbf{a}+\mathbf{b}][\mathbf{b} \times \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]} \\ & =\frac{[[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{c}[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{a}[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{b}]}{2[\mathbf{a} \mathbf{b} \mathbf{c}][\mathbf{a} \mathbf{b}]^2} \\ & =\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]^3[\mathbf{a} \mathbf{b} \mathbf{c}]}{2[\mathbf{a} \mathbf{b} \mathbf{c}]^3}=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{2}=1 \\ & \left\lceil\because[\mathbf{a} \mathbf{b} \mathbf{c}]=\left|\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array}\right|=2\right\rfloor \\ & \end{aligned} $$

Hence, the correct answer is option: 'B'.

Solution

Relevant Questions