 # 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 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.

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

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:

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

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

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

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

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