Question

A unit vector \(\bar{u}\) makes an angle of \(45^{\circ }\) with the X- axis. If \(\bar{u}+\hat{j}+\hat{k}\) is also a unit vector, then the sum of the projections of \(\bar{u}\) on the coordinate axes is:

• A. \(\sqrt{2}-1\)
• B. \(\sqrt{2}+1\)
• C. \(\frac{1}{\sqrt{2}}-1\)
• D. \(\frac{1}{\sqrt{2}}+1\)

Hint:

The projections are the components of the unit vector. Use the unit conditions on u and on u plus j plus k.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The projections of \(\bar u\) on the axes are its components. Let \(\bar u=(u_1,u_2,u_3)\).
The angle with the X-axis is \(45^{\circ}\), so \(u_1=\cos45^{\circ}=\dfrac1{\sqrt2}\).

Step 2: Unit Vector Condition
\(u_1^{2}+u_2^{2}+u_3^{2}=1\), so \(u_2^{2}+u_3^{2}=\dfrac12\).

Step 3: Second Condition
\(\bar u+\hat j+\hat k=\left(u_1,u_2+1,u_3+1\right)\) is a unit vector:
\(\dfrac12+(u_2+1)^{2}+(u_3+1)^{2}=1\).
Expanding: \(\dfrac12+\dfrac12+2(u_2+u_3)+2=1\), so \(u_2+u_3=-1\).

Step 4: Calculation
Sum of projections \(=\dfrac1{\sqrt2}+u_2+u_3=\dfrac1{\sqrt2}-1\).

Final Answer:
The sum of the projections is 1 over root 2 minus 1. \[ \boxed{\text{(C) }\dfrac{1}{\sqrt2}-1} \]