Question

If the vector \( \vec{v} = \vec{i} - 7\vec{j} + 2\vec{k} \) is along the internal bisector of the angle between the vectors \( \vec{a} \) and \( \vec{b} = -2\vec{i} - \vec{j} + 2\vec{k} \) and the unit vector along \( \vec{a} \) is \( \hat{a} = x\vec{i} + y\vec{j} + z\vec{k} \) then \( x = \)

• A. 0
• B. \( \frac{7}{9} \)
• C. \( \frac{1}{9} \)
• D. \( \frac{5}{3} \) Correct Answer

Hint:

A vector along the internal bisector of the angle between \( \vec{a} \) and \( \vec{b} \) is proportional to \( \hat{a} + \hat{b} \), i.e., \( k(\frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|}) \). If \( \vec{v} \) is such a vector, then \( \vec{v} = \lambda (\hat{a} + \hat{b}) \). Equate components and use the property that \( \hat{a} \) is a unit vector (\( |\hat{a}|=1 \)).

Answer 1

The Correct Option is B

Step 1: Recall the formula for a vector along the internal angle bisector.
A vector along the internal bisector of the angle between vectors \( \vec{a} \) and \( \vec{b} \) is given by \( \lambda \left( \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|} \right) = \lambda (\hat{a} + \hat{b}) \), where \( \lambda>0 \).

Step 2: Calculate \( |\vec{b}| \) and \( \hat{b} \).
\( \vec{b} = -2\vec{i} - \vec{j} + 2\vec{k} \).
\( |\vec{b}| = \sqrt{(-2)^2 + (-1)^2 + (2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3 \).
\( \hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{-2\vec{i} - \vec{j} + 2\vec{k}}{3} = -\frac{2}{3}\vec{i} - \frac{1}{3}\vec{j} + \frac{2}{3}\vec{k} \).

Step 3: Set up the equation for the bisector vector.
We are given that \( \vec{v} = \vec{i} - 7\vec{j} + 2\vec{k} \) is along the bisector.
So, \( \vec{v} = \lambda (\hat{a} + \hat{b}) \) for some \( \lambda>0 \).
Let \( \hat{a} = x\vec{i} + y\vec{j} + z\vec{k} \).
Then \( \hat{a} + \hat{b} = (x - \frac{2}{3})\vec{i} + (y - \frac{1}{3})\vec{j} + (z + \frac{2}{3})\vec{k} \).
So, \( \vec{i} - 7\vec{j} + 2\vec{k} = \lambda \left( (x - \frac{2}{3})\vec{i} + (y - \frac{1}{3})\vec{j} + (z + \frac{2}{3})\vec{k} \right) \).

Step 4: Equate the components.
\( 1 = \lambda (x - \frac{2}{3}) \) (Eq.
1) \( -7 = \lambda (y - \frac{1}{3}) \) (Eq.
2) \( 2 = \lambda (z + \frac{2}{3}) \) (Eq.
3) Also, since \( \hat{a} \) is a unit vector, \( x^2+y^2+z^2=1 \) (Eq.
4).

Step 5: Solve for \(x, y, z, \lambda\).
This seems complex with 4 equations.
Alternative thinking: \( \vec{v} \) is parallel to \( \hat{a} + \hat{b} \).
\( |\vec{v}| = \sqrt{1^2 + (-7)^2 + 2^2} = \sqrt{1+49+4} = \sqrt{54} = 3\sqrt{6} \).
The unit vector along \( \vec{v} \) is \( \hat{v} = \frac{1}{3\sqrt{6}}(\vec{i} - 7\vec{j} + 2\vec{k}) \).
Since \( \vec{v} \) is along \( \hat{a} + \hat{b} \), then \( \hat{v} \) must be the unit vector of \( \hat{a} + \hat{b} \).
So, \( \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|} = \frac{\vec{i} - 7\vec{j} + 2\vec{k}}{3\sqrt{6}} \).
This implies \( \hat{a} + \hat{b} = k(\vec{i} - 7\vec{j} + 2\vec{k}) \) where \( k = \frac{|\hat{a} + \hat{b}|}{3\sqrt{6}} \).
This is essentially the same as before, with \( \lambda = 1/k \).
From (1), (2), (3): \( x - \frac{2}{3} = \frac{1}{\lambda} \) \( y - \frac{1}{3} = \frac{-7}{\lambda} \) \( z + \frac{2}{3} = \frac{2}{\lambda} \) So, \( x = \frac{1}{\lambda} + \frac{2}{3} \), \( y = \frac{-7}{\lambda} + \frac{1}{3} \), \( z = \frac{2}{\lambda} - \frac{2}{3} \).
Substitute into \( x^2+y^2+z^2=1 \).
\( \left(\frac{1}{\lambda} + \frac{2}{3}\right)^2 + \left(\frac{-7}{\lambda} + \frac{1}{3}\right)^2 + \left(\frac{2}{\lambda} - \frac{2}{3}\right)^2 = 1 \) \( \frac{1}{\lambda^2} + \frac{4}{3\lambda} + \frac{4}{9} + \frac{49}{\lambda^2} - \frac{14}{3\lambda} + \frac{1}{9} + \frac{4}{\lambda^2} - \frac{8}{3\lambda} + \frac{4}{9} = 1 \) Collect terms: \( \frac{1+49+4}{\lambda^2} + \frac{4-14-8}{3\lambda} + \frac{4+1+4}{9} = 1 \) \( \frac{54}{\lambda^2} + \frac{-18}{3\lambda} + \frac{9}{9} = 1 \) \( \frac{54}{\lambda^2} - \frac{6}{\lambda} + 1 = 1 \) \( \frac{54}{\lambda^2} - \frac{6}{\lambda} = 0 \) Since \( \lambda \ne 0 \), multiply by \( \lambda^2 \): \( 54 - 6\lambda = 0 \implies 6\lambda = 54 \implies \lambda = 9 \).
Now find \(x\): \( x = \frac{1}{\lambda} + \frac{2}{3} = \frac{1}{9} + \frac{2}{3} = \frac{1}{9} + \frac{6}{9} = \frac{7}{9} \).
This matches option (2).