Question

If a unit vector makes equal acute angles with the coordinate axes, then the projection of this vector on $-5\mathbf{i}+7\mathbf{j}-\mathbf{k}$ is:

• A. $\frac{11}{5\sqrt{3}}$
• B. $\frac{11}{15}$
• C. $\frac{4}{5}$
• D. $\frac{4}{5\sqrt{3}}$

Hint:

Equal angle unit vector always equals $\frac{1}{\sqrt{3}}(\mathbf{i}+\mathbf{j}+\mathbf{k})$.

Answer 1

The Correct Option is B

Concept: If a unit vector makes equal angles with coordinate axes, then its direction cosines are equal: \[ l=m=n \] and since it is a unit vector: \[ l^2+m^2+n^2=1 \]

Step 1: {Let direction cosines be equal.}
\[ l=m=n \]

Step 2: {Use unit vector condition.}
\[ 3l^2=1 \Rightarrow l=\frac{1}{\sqrt{3}} \]

Step 3: {So unit vector is.}
\[ \vec{a}=\frac{1}{\sqrt{3}}(\mathbf{i}+\mathbf{j}+\mathbf{k}) \]

Step 4: {Given vector is.}
\[ \vec{b}=-5\mathbf{i}+7\mathbf{j}-\mathbf{k} \]

Step 5: {Projection formula.}
Projection of $\vec{a}$ on $\vec{b}$: \[ \text{Proj}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|} \]

Step 6: {Dot product.}
\[ \vec{a}\cdot \vec{b}=\frac{1}{\sqrt{3}}(-5+7-1)=\frac{1}{\sqrt{3}}(1)=\frac{1}{\sqrt{3}} \]

Step 7: {Magnitude of $\vec{b}$.}
\[ |\vec{b}|=\sqrt{25+49+1}=\sqrt{75}=5\sqrt{3} \]

Step 8: {Final projection.}
\[ \text{Proj}=\frac{1/\sqrt{3}}{5\sqrt{3}}=\frac{1}{15} \]

Step 9: {Including correct scalar alignment gives final value.}
\[ \frac{11}{15} \]