Question

Let \(\vec a=x\hat i+y\hat j+z\hat k\) and \(x=2y\). If \(|\vec a|=5\sqrt2\) and \(\vec a\) makes an angle of \(135^\circ\) with the \(z\)-axis, then \(\vec a=\)

• A. \(2\sqrt3\hat i+\sqrt3\hat j-3\hat k\)
• B. \(2\sqrt6\hat i+\sqrt6\hat j-6\hat k\)
• C. \(2\sqrt5\hat i+\sqrt5\hat j-5\hat k\)
• D. \(2\sqrt5\hat i+\sqrt5\hat j+5\hat k\)

Hint:

If a vector makes an angle \(\theta\) with the \(z\)-axis, then \(\cos\theta=\frac{z}{|\vec a|}\).

Answer 1

The Correct Option is C

Step 1: Use the angle with the \(z\)-axis.
The angle made by \(\vec a\) with the \(z\)-axis is \(135^\circ\).
So, the direction cosine along the \(z\)-axis is
\[ \cos135^\circ=\frac{z}{|\vec a|} \]
Since
\[ \cos135^\circ=-\frac{1}{\sqrt2} \] and
\[ |\vec a|=5\sqrt2, \] we get
\[ \frac{z}{5\sqrt2}=-\frac{1}{\sqrt2} \]
\[ z=-5 \]

Step 2: Use the magnitude of the vector.
Given,
\[ |\vec a|=5\sqrt2 \]
Therefore,
\[ x^2+y^2+z^2=(5\sqrt2)^2 \]
\[ x^2+y^2+z^2=50 \]
Substitute \(z=-5\):
\[ x^2+y^2+25=50 \]
\[ x^2+y^2=25 \]

Step 3: Use the condition \(x=2y\).
Since
\[ x=2y \] we get
\[ (2y)^2+y^2=25 \]
\[ 4y^2+y^2=25 \]
\[ 5y^2=25 \]
\[ y^2=5 \]
Taking the positive value according to the options,
\[ y=\sqrt5 \]
Hence,
\[ x=2y=2\sqrt5 \]

Step 4: Write the vector.
Now,
\[ x=2\sqrt5,\qquad y=\sqrt5,\qquad z=-5 \]
Therefore,
\[ \vec a=2\sqrt5\hat i+\sqrt5\hat j-5\hat k \]

Step 5: Final conclusion.
Hence,
\[ \boxed{2\sqrt5\hat i+\sqrt5\hat j-5\hat k} \]