If a - i - j - and b - j - k - find the projection of a - on b -

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Standard XII

Mathematics

Definite Integral as Limit of Sum

If a →= i ∧- ...

Question

If $\vec{a} = \hat{i} - \hat{j} and \vec{b} = - \hat{j} + \hat{k},$ find the projection of $\vec{a} on \vec{b}$ .

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Solution

$We have \vec{a} = \hat{i} - \hat{j} a n d \vec{b} = - \hat{j} + \hat{k} The projection of \vec{a} on \vec{b} is \frac{\vec{a} . \vec{b}}{|\vec{b}|} = \frac{(\hat{i} - \hat{j}) . (- \hat{j} + \hat{k})}{|- \hat{j} + \hat{k}|} = \frac{0 + 1 + 0}{\sqrt{1 + 1}} = \frac{1}{\sqrt{2}}$

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If a →=i^-j^ and b →=-j^+k^, find the projection of a →on b →.

We havea→=i^-j^ and b→=-j^+k^The projection of a→ on b→ isa→. b→b→=i^-j^.-j^+k^-j^+k^=0+1+01+1=12

If $\vec{a} = \hat{i} - \hat{j} and \vec{b} = - \hat{j} + \hat{k},$ find the projection of $\vec{a} on \vec{b}$ .

$We have \vec{a} = \hat{i} - \hat{j} a n d \vec{b} = - \hat{j} + \hat{k} The projection of \vec{a} on \vec{b} is \frac{\vec{a} . \vec{b}}{|\vec{b}|} = \frac{(\hat{i} - \hat{j}) . (- \hat{j} + \hat{k})}{|- \hat{j} + \hat{k}|} = \frac{0 + 1 + 0}{\sqrt{1 + 1}} = \frac{1}{\sqrt{2}}$