# Find the scalar and vector projections of b onto a. a=i+j+k, b=i−j+k

The aim of this question is to find the Scalar and Vector Projection of the given two vectors.

The basic concept behind this article is the understanding of Scalar and Vector Projections of vector quantities and how to calculate them.

The Scalar Projection of one vector $\vec{a}$ onto another vector $\vec{b}$ is expressed as the length of vector $\vec{a}$ being projected on the length of vector $\vec{b}$. It is calculated by taking the dot product of both vector $\vec{a}$ and vector $\vec{b}$ and then dividing it by the modular value of the vector on which it is being projected.

$Scalar\ Projection\ S_{b\rightarrow a}\ =\ \frac{\vec{a}\ .\vec{b}}{\left|\vec{b}\right|}$

The Vector Projection of one vector $\vec{a}$ onto another vector $\vec{b}$ is expressed as the shadow or orthogonal projection of vector $\vec{a}$ on a straight line that is parallel to vector $\vec{b}$. It is calculated by multiplying the Scalar Projection of both vectors by the unitary vector on which it is being projected.

$Vector\ Projection\ V_{a\rightarrow b}=\frac{\vec{a}\ .\vec{b}}{\left|\vec{b}\right|^2}(\vec{b})$

## Expert Answer

Given that:

Vector $\vec{a}=\hat{i}+\hat{j}+\hat{k}$

Vector $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

We are given that vector $\vec{b}$ is projected on vector $\vec{a}$.

The Scalar Projection of vector $\vec{b}$ projected on vector $\vec{a}$ will be calculated as follows:

$Scalar\ Projection\ S_{b\rightarrow a}\ =\ \frac{\vec{a}\ .\vec{b}}{\left|\vec{a}\right|}$

Substituting the given values in the above equation:

$S_{b\rightarrow a}=\frac{(\hat{i}+\hat{j}+\hat{k})\ .(\hat{i}-\hat{j}+\hat{k})}{\left|\hat{i}+\hat{j}+\hat{k}\right|}$

We know that:

$\left|a\hat{i}+b\hat{j}+c\widehat{k}\right|=\sqrt{a^2+b^2+c^2}$

Using this concept:

$S_{b\rightarrow a}=\frac{(\hat{i}+\hat{j}+\hat{k})\ .(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^2+1^2}}$

$S_{b\rightarrow a}=\frac{1^2-1^2+1^2}{\sqrt{1^2+1^2+1^2}}$

$S_{b\rightarrow a}=\frac{1-1+1}{\sqrt{1+1+1}}$

$Scalar\ Projection\ S_{b\rightarrow a}=\frac{1}{\sqrt3}$

The Vector Projection of vector $\vec{b}$ projected on vector $\vec{a}$ will be calculated as follows:

$Vector\ Projection\ V_{b\rightarrow a}=\frac{\vec{a}\ .\vec{b}}{\left|\vec{a}\right|^2}(\vec{a})$

Substituting the given values in the above equation:

$V_{b\rightarrow a}=\frac{(\hat{i}+\hat{j}+\hat{k})\ .(\hat{i}-\hat{j}+\hat{k})}{\left|\hat{i}+\hat{j}+\hat{k}\right|^2}\times(\hat{i}+\hat{j}+\hat{k})$

$V_{b\rightarrow a}=\frac{1^2-1^2+1^2}{{(\sqrt{1^2+1^2+1^2})}^2}\times(\hat{i}+\hat{j}+\hat{k})$

$V_{b\rightarrow a}=\frac{1-1+1}{{(\sqrt{1+1+1})}^2}\times(\hat{i}+\hat{j}+\hat{k})$

$V_{b\rightarrow a}=\frac{1}{3}\times(\hat{i}+\hat{j}+\hat{k})$

${Vector\ Projection\ V}_{b\rightarrow a}=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$

## Numerical Result

The Scalar Projection of vector $\vec{b}$ projected on vector $\vec{a}$ is as follows:

$Scalar\ Projection\ S_{b\rightarrow a}\ =\ \frac{1}{\sqrt3}$

The Vector Projection of vector $\vec{b}$ projected on vector $\vec{a}$ is as follows:

${Vector\ Projection\ V}_{b\rightarrow a}\ =\ \frac{1}{3}\ (\hat{i}\ +\ \hat{j}\ +\ \hat{k})$

## Example

For the given vector $\vec{a}$ and vector $\vec{b}$, calculate the Scalar and Vector Projection of vector $\vec{b}$ onto vector $\vec{a}$.

Vector $\vec{a}\ =\ 3\widehat{i}\ -\ \hat{j}\ +\ 4\hat{k}$

Vector $\vec{b}\ =\widehat{j}\ +\ \dfrac{1}{2}\hat{k}$

Solution

The Scalar Projection of vector $\vec{b}$ projected on vector $\vec{a}$ will be calculated as follows:

$Scalar\ Projection\ S_{b\rightarrow a}\ =\ \frac{\vec{a}\ .\vec{b}}{\left|\vec{a}\right|}$

Substituting the given values in the above equation:

$S_{b\rightarrow a}\ =\ \frac{(3\hat{i}\ -\ \hat{j}\ +\ 4\hat{k})\ .(0\hat{i}\ +\ \hat{j}\ +\ \dfrac{1}{2}\hat{k})}{\left|3\hat{i}\ -\ \hat{j}+\ 4\hat{k}\right|}$

$S_{b\rightarrow a}\ =\ \frac{(3)\ (0)\ +\ (-1)\ (1)\ +\ (4)\ \left(\dfrac{1}{2}\right)}{\sqrt{{(3)}^2+{\ \ (-1)}^2\ +{\ (4)}^2}}$

$S_{b\rightarrow a}\ =\frac{0\ -\ 1\ \ +2}{\ \sqrt{9+\ 1\ \ +\ 16}}$

$S_{b\rightarrow a}=\ \ \frac{1}{\sqrt{26}}$

$Scalar\ Projection\ \ S_{b\rightarrow a}\ =\ \frac{1}{\sqrt6}$

The Vector Projection of vector $\vec{b}$ projected on vector $\vec{a}$ will be calculated as follows:

$Vector\ Projection\ {\ V}_{b\rightarrow a}\ =\ \frac{\vec{a}\ .\vec{b}}{\left|\vec{a}\right|^2}\ (\vec{a})$

Substituting the given values in the above equation:

$V_{b\rightarrow a}\ =\ \frac{(3\hat{i}\ -\ \hat{j}\ +\ 4\hat{k})\ .(0\hat{i}\ +\ \hat{j}+\ \ \dfrac{1}{2}\hat{k})}{\left|3\hat{i}\ -\ \hat{j}\ +\ 4\hat{k}\right|^2}\ \times\ (3\hat{i}-\ \ \hat{j}\ +\ 4\hat{k})$

$V_{b\rightarrow a}\ =\ \frac{(3)\ (0)\ +\ (-1)\ (1)\ +\ (4)\ \left(\dfrac{1}{2}\right)}{{(\sqrt{{(3)}^2\ +\ {(-1)}^2\ +{\ (4)}^2})}^2}\ \times\ (3\hat{i}\ -\ \hat{j}\ +\ 4\hat{k})$

$V_{b\rightarrow a}\ =\ \frac{0\ -\ 1\ +\ 2}{{(\sqrt{26})}^2}\ \times\ (3\hat{i}\ -\ \hat{j}\ +\ 4\hat{k})$

$V_{b\rightarrow a}\ =\frac{1}{\ 26}\ \times\ (3\hat{i}\ -\ \hat{j}\ +\ 4\hat{k})$

${Vector\ Projection\ V}_{b\rightarrow a}\ =\ \frac{1}{3}\ (3\hat{i}\ -\ \hat{j}\ +\ 4\hat{k})$

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