Question

Solve the given expression:

Answer 1

Answer:

The given vectors cannot be orthogonal

Step-by-step explanation:

Let

$\vec{x}=a\hat{i}+3\hat{j}+2\hat{k}$

$\vec{y}=-a\hat{i}+\hat{j}-2\hat{k}$

$\text{suppose }\vec{x}\text{ and }\vec{y}\:\text{ are orthogonal}$

$\text{Then }\vec{x}.\vec{y}=0$

$-a^2+3-4=0$

$-a^2-1=0$

$a^2+1=0$

$a=\pm\sqrt{-1}$

$\text{a is imaginary}$

$\implies\text{ a cannot be real }$

Hence for all real values of a, $\vec{x}\:and\:\vec{y}$ are cannot orthogonal

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

If possible let the given two vectors are orthogonal for some real values of a

Then

$\sf{( \: a \hat{i} + 3 \hat{j} + 2\hat{k} \: )} .( \: - a \hat{i} + \hat{j} - 2\hat{k} \: ) = 0$

$\implies \sf{ - {a}^{2} + 3 - 4} = 0$

$\implies \sf{ {a}^{2} } = - 1$

Which is absurd

Since there exists no real number a such that

$\sf{ {a}^{2} } = - 1$

RESULT

Hence the the vectors

$\sf{( \: a \hat{i} + 3 \hat{j} + 2\hat{k} \: )} \: \: and \: \: ( \: - a \hat{i} + \hat{j} - 2\hat{k} \: )$

Can not be orthogonal

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LEARN MORE FROM BRAINLY

Divergence of r / r3 is

(a) zero at the origin

(b) zero everywhere

(c) zero everywhere except the origin

(d) nonzero

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