Question

The vector 6î+3j^-ak^ is perpendicular to 3î+j^+3k^. The value of a is

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Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

The vector 6î+3j^-ak^ is perpendicular to 3î+j^+3k^. The value of a is

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{\pink{Given}}}}}$

• two vector 6î+3j^-ak^ and 3î+j^+3k^
• both are perpendicular

$\Large{\underline{\mathfrak{\bf{\pink{Find}}}}}$

• Value of " a "

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

we know,

If two vector are in perpendicular ,

So, their scaler product always will be zero (0) .

Let,

$\mapsto\sf{\:\vec{A}\:=\:6\^{i}+3\^{j}-a\^{k}}$

$\mapsto\sf{\:\vec{B}\:=\:3\^{i}+\^{j}+3\^{k}}$

$\Large{\underline{\mathfrak{\bf{\orange{Scaler\:product}}}}}$

$\mapsto\sf{\:\vec{A}.\vec{B}\:=\:(6\^{i}+3\^{j}-a\^{k}).(3\^{i}+\^{j}+3\^{k})\:=\:0}}} \\ \\ \mapsto\sf{\:\vec{A}.\vec{B}\:=\:(6\^{i}).(3\^{i})+(3\^{j}).(\^{j})-(a\^{k}).(3\^{k})\:=\:0}}$

$\mapsto\sf{\:\vec{A}.\vec{B}\:=\:(18+3-3a)\:=\:0}} \\ \\ \mapsto\sf{\:3a\:=\:21}} \\ \\ \mapsto\sf{\:a\:=\:\cancel{\dfrac{21}{3}}}} \\ \\ \mapsto\sf{\red{\:a\:=\:7}}$

$\Large{\underline{\mathfrak{\bf{\orangr{Thus}}}}}$

• Value of a = 7

★Important Scaler Product

• $\sf{\:\^{i}.\^{i}\:=\:1}$
• $\sf{\:\^{j}.\^{j}\:=\:1}$
• $\sf{\:\^{k}.\^{k}\:=\:1}$
• $\sf{\:\^{i}.\^{j}\:=\:0}$
• $\sf{\:\^{j}.\^{k}\:=\:0}$
• $\sf{\:\^{k}.\^{i}\:=\:0}$

Answer 2

the answer of this question is a=7