Question

A vector a=αi+2j+βk (α,β∈R) lies in the plane of the vectors b=i+j and vector c=i-j+4k. If vector a bisects the angle between vector b and vector c then

• A. a.k+4=0
• B. a.k+2=0
• C. a.i+1=0
• D. a.i+3=0

Answer 1

Answer: (B)

a.k+2=0

Answer 2

1. Vector a in the plane of b and c: This condition implies that the scalar triple product [a b c] is zero, i.e., a . (b x c) = 0.

   • Calculate b x c: $b \times c = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 1 & -1 & 4 \end{vmatrix} = \mathbf{i}(4-0) - \mathbf{j}(4-0) + \mathbf{k}(-1-1) = 4\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}$
   • Calculate a . (b x c): $(\alpha\mathbf{i} + 2\mathbf{j} + \beta\mathbf{k}) \cdot (4\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}) = 4\alpha - 8 - 2\beta$
   • Setting the dot product to zero: $4\alpha - 8 - 2\beta = 0$, which simplifies to $2\alpha - \beta = 4$ (Equation 1).

2. Vector a bisects the angle between b and c: This means a is parallel to the sum or difference of the unit vectors of b and c.

   • Calculate magnitudes: $|b| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$ $|c| = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$
   • Unit vectors: $u_b = \frac{b}{|b|} = \frac{\mathbf{i} + \mathbf{j}}{\sqrt{2}}$ $u_c = \frac{c}{|c|} = \frac{\mathbf{i} - \mathbf{j} + 4\mathbf{k}}{3\sqrt{2}}$

3. Case 1: a is parallel to the internal angle bisector ($u_b + u_c$)

   • $u_b + u_c = \frac{\mathbf{i} + \mathbf{j}}{\sqrt{2}} + \frac{\mathbf{i} - \mathbf{j} + 4\mathbf{k}}{3\sqrt{2}} = \frac{3(\mathbf{i} + \mathbf{j}) + (\mathbf{i} - \mathbf{j} + 4\mathbf{k})}{3\sqrt{2}} = \frac{4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}}{3\sqrt{2}}$
   • So, $a$ is parallel to $4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}$. Let $a = K(4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})$.
   • Comparing the coefficient of $\mathbf{j}$ in $a = \alpha\mathbf{i} + 2\mathbf{j} + \beta\mathbf{k}$ and $a = K(4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})$, we get $2 = 2K$, so $K=1$.
   • This gives $a = 4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}$, implying $\alpha = 4$ and $\beta = 4$.
   • Check Equation 1: $2(4) - 4 = 8 - 4 = 4$. This is satisfied.
   • Check options:
     • (a) $a \cdot \mathbf{k} + 4 = 4 + 4 = 8 \neq 0$
     • (b) $a \cdot \mathbf{k} + 2 = 4 + 2 = 6 \neq 0$
     • (c) $a \cdot \mathbf{i} + 1 = 4 + 1 = 5 \neq 0$
     • (d) $a \cdot \mathbf{i} + 3 = 4 + 3 = 7 \neq 0$
   • None of the options are satisfied.

4. Case 2: a is parallel to the external angle bisector ($u_b - u_c$)

   • $u_b - u_c = \frac{\mathbf{i} + \mathbf{j}}{\sqrt{2}} - \frac{\mathbf{i} - \mathbf{j} + 4\mathbf{k}}{3\sqrt{2}} = \frac{3(\mathbf{i} + \mathbf{j}) - (\mathbf{i} - \mathbf{j} + 4\mathbf{k})}{3\sqrt{2}} = \frac{2\mathbf{i} + 4\mathbf{j} - 4\mathbf{k}}{3\sqrt{2}}$
   • So, $a$ is parallel to $2\mathbf{i} + 4\mathbf{j} - 4\mathbf{k}$. Let $a = K'(2\mathbf{i} + 4\mathbf{j} - 4\mathbf{k})$.
   • Comparing the coefficient of $\mathbf{j}$ in $a = \alpha\mathbf{i} + 2\mathbf{j} + \beta\mathbf{k}$ and $a = K'(2\mathbf{i} + 4\mathbf{j} - 4\mathbf{k})$, we get $2 = 4K'$, so $K' = 1/2$.
   • This gives $a = \frac{1}{2}(2\mathbf{i} + 4\mathbf{j} - 4\mathbf{k}) = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$. So, $\alpha = 1$ and $\beta = -2$.
   • Check Equation 1: $2(1) - (-2) = 2 + 2 = 4$. This is satisfied.
   • Check options:
     • (a) $a \cdot \mathbf{k} + 4 = -2 + 4 = 2 \neq 0$
     • (b) $a \cdot \mathbf{k} + 2 = -2 + 2 = 0$. This option is correct.
     • (c) $a \cdot \mathbf{i} + 1 = 1 + 1 = 2 \neq 0$
     • (d) $a \cdot \mathbf{i} + 3 = 1 + 3 = 4 \neq 0$
   • Option (b) is satisfied.