Question

The vectors \(3\vec a-5\vec b\) and \(2\vec a+\vec b\) are mutually perpendicular and the vectors \(\vec a+4\vec b\) and \(-\vec a+\vec b\) are also mutually perpendicular. Then the acute angle between \(\vec a\) and \(\vec b\) is

• A. \(\cos^{-1}\left(\dfrac{19}{5\sqrt{43}}\right)\)
• B. \(\cos^{-1}\left(\dfrac{9}{5\sqrt{43}}\right)\)
• C. \(\pi-\cos^{-1}\left(\dfrac{19}{5\sqrt{43}}\right)\)
• D. \(\pi-\cos^{-1}\left(\dfrac{9}{5\sqrt{43}}\right)\)

Hint:

Whenever vectors are perpendicular, immediately use the dot product equal to zero. Then express \(\vec a\cdot\vec b\) as \(|\vec a||\vec b|\cos\theta\).

Answer 1

The Correct Option is A

Step 1: Use the condition of perpendicular vectors.
If two vectors are perpendicular, then their dot product is zero.
Given,
\[ (3\vec a-5\vec b)\cdot(2\vec a+\vec b)=0 \]
Expanding,
\[ 6|\vec a|^2+3\vec a\cdot\vec b-10\vec a\cdot\vec b-5|\vec b|^2=0 \]
\[ 6|\vec a|^2-7\vec a\cdot\vec b-5|\vec b|^2=0 \]
\[ 6|\vec a|^2-5|\vec b|^2=7\vec a\cdot\vec b \tag{1} \]

Step 2: Use the second perpendicular condition.
Also,
\[ (\vec a+4\vec b)\cdot(-\vec a+\vec b)=0 \]
Expanding,
\[ -|\vec a|^2+\vec a\cdot\vec b-4\vec a\cdot\vec b+4|\vec b|^2=0 \]
\[ -|\vec a|^2-3\vec a\cdot\vec b+4|\vec b|^2=0 \]
\[ 4|\vec b|^2-|\vec a|^2=3\vec a\cdot\vec b \tag{2} \]

Step 3: Introduce standard notation.
Let
\[ |\vec a|=A,\qquad |\vec b|=B \] and
\[ \vec a\cdot\vec b=AB\cos\theta \]
Then equations (1) and (2) become
\[ 6A^2-5B^2=7AB\cos\theta \tag{3} \]
and
\[ 4B^2-A^2=3AB\cos\theta \tag{4} \]

Step 4: Find the ratio \(A:B\).
Multiply equation (4) by \(7\):
\[ 28B^2-7A^2=21AB\cos\theta \]
Multiply equation (3) by \(3\):
\[ 18A^2-15B^2=21AB\cos\theta \]
Equating,
\[ 28B^2-7A^2=18A^2-15B^2 \]
\[ 43B^2=25A^2 \]
\[ \frac{A}{B}=\frac{\sqrt{43}}{5} \tag{5} \]

Step 5: Find \(\cos\theta\).
Using equation (4),
\[ 3AB\cos\theta=4B^2-A^2 \]
Substitute \(A^2=\frac{43}{25}B^2\):
\[ 3AB\cos\theta = 4B^2-\frac{43}{25}B^2 \]
\[ = \frac{100-43}{25}B^2 \]
\[ = \frac{57}{25}B^2 \]
Therefore,
\[ \cos\theta = \frac{57B^2}{75AB} \]
\[ = \frac{19B}{25A} \]
Using \(\dfrac{A}{B}=\dfrac{\sqrt{43}}{5}\),
\[ \frac{B}{A}=\frac{5}{\sqrt{43}} \]
Hence,
\[ \cos\theta = \frac{19}{25}\cdot\frac{5}{\sqrt{43}} \]
\[ = \frac{19}{5\sqrt{43}} \]
Since the acute angle is required,
\[ \theta= \cos^{-1}\left(\frac{19}{5\sqrt{43}}\right) \]

Step 6: Final conclusion.
Hence,
\[ \boxed{ \cos^{-1}\left(\frac{19}{5\sqrt{43}}\right) } \]