 # How do you find the scalar and vector projections of Mary Buchanan 2021-12-15 Answered
How do you find the scalar and vector projections of b onto a given $a=i+j+k$, $b=i-j+k$?
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We have two vectors ${a}^{\to }$ and ${b}^{\to }$ and we have to find the scalar and vector projection of b onto a
${a}^{\to }=i+j+k$ and ${b}^{\to }=i-j+k$
The scalar projection of ${b}^{\to }$ onto ${a}^{\to }=\frac{{b}^{\to }×{a}^{\to }}{|{a}^{\to }|}$
$=\frac{\left(i+j+k\right)\left(i-j+k\right)}{\sqrt{{1}^{2}+{1}^{1}+{1}^{2}}}$
$=\frac{{1}^{2}-{1}^{2}+{1}^{2}}{\sqrt{3}}$
$=\frac{1}{\sqrt{3}}$
The vector projection of ${b}^{\to }$ onto ${a}^{\to }=\frac{{b}^{\to }×{a}^{\to }}{{|{a}^{\to }|}^{2}}×\left(i+j+k\right)$
$=\frac{\left(i+j+k\right)\left(i-j+k\right)}{{\left(\sqrt{{1}^{2}+{1}^{1}+{1}^{2}}\right)}^{2}}×\left(i+j+k\right)$
$=\frac{{1}^{2}-{1}^{2}+{1}^{2}}{3}×\left(i+j+k\right)$
$=\frac{1}{3}\left(i+j+k\right)$

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