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Question: If \(\overset{\rightarrow}{a}\)= \(\overset{\hat{}}{i}\)+ \(\overset{\hat{}}{j}\) + \(\overset{\hat{...

If a→\overset{\rightarrow}{a}= i^\overset{\hat{}}{i}+ j^\overset{\hat{}}{j} + k^\overset{\hat{}}{k}, b→\overset{\rightarrow}{b}= i^\overset{\hat{}}{i} –j^\overset{\hat{}}{j} + k^\overset{\hat{}}{k}, c→\overset{\rightarrow}{c}= i^\overset{\hat{}}{i}+ 2j^\overset{\hat{}}{j} – k^\overset{\hat{}}{k}, then the value of ∣aˉ.aˉaˉ.bˉaˉ.cˉbˉ.aˉbˉ.bˉbˉ.cˉcˉ.aˉcˉ.bˉcˉ.cˉ∣\left| \begin{matrix} \bar{a}.\bar{a} & \bar{a}.\bar{b} & \bar{a}.\bar{c} \\ \bar{b}.\bar{a} & \bar{b}.\bar{b} & \bar{b}.\bar{c} \\ \bar{c}.\bar{a} & \bar{c}.\bar{b} & \bar{c}.\bar{c} \end{matrix} \right|is

A

16

B

18

C

19

D

20

Answer

16

Explanation

Solution

a→\overset{\rightarrow}{a}.a→\overset{\rightarrow}{a}= (i^\overset{\hat{}}{i} + j^\overset{\hat{}}{j} + k^\overset{\hat{}}{k}). (i^\overset{\hat{}}{i} + j^\overset{\hat{}}{j} + k^\overset{\hat{}}{k})

= 1 + 1 + 1 = 3

b→\overset{\rightarrow}{b}.b→\overset{\rightarrow}{b}= 1 + 1 + 1 = 3

c→\overset{\rightarrow}{c}.c→\overset{\rightarrow}{c}= 1 + 4 + 1 = 6

a→\overset{\rightarrow}{a}.b→\overset{\rightarrow}{b}= 1 – 1 + 1 = 1=b→\overset{\rightarrow}{b}.a→\overset{\rightarrow}{a}

b→\overset{\rightarrow}{b}.c→\overset{\rightarrow}{c}= 1 – 2 – 1 = –2 =c→\overset{\rightarrow}{c}. b→\overset{\rightarrow}{b}

c→\overset{\rightarrow}{c}.a→\overset{\rightarrow}{a}= 1 + 2 – 1 = 2 =a→\overset{\rightarrow}{a} . c→\overset{\rightarrow}{c}

Value =∣31213−22−26∣\left| \begin{matrix} 3 & 1 & 2 \\ 1 & 3 & - 2 \\ 2 & - 2 & 6 \end{matrix} \right| = 16