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. Coplanarity Condition: Since vector $a$ lies in the plane of vectors $b$ and $c$, they are coplanar. This means their scalar triple product is zero: $a \cdot (b \times c) = 0$.

   • Calculate $b \times c$: $b \times c = (i + j) \times (i - j + 4k) = \begin{vmatrix} i & j & k \\ 1 & 1 & 0 \\ 1 & -1 & 4 \end{vmatrix} = i(4-0) - j(4-0) + k(-1-1) = 4i - 4j - 2k$.
   • Calculate $a \cdot (b \times c)$: $(\alpha i + 2j + \beta k) \cdot (4i - 4j - 2k) = 4\alpha - 8 - 2\beta$.
   • Set the scalar triple product to zero: $4\alpha - 8 - 2\beta = 0 \implies 2\alpha - \beta = 4$ (Equation 1).

2. Angle Bisector Condition: Vector $a$ bisects the angle between $b$ and $c$. This implies $a$ is parallel to either the sum or the difference of the unit vectors of $b$ and $c$.

   • Calculate unit vectors: $|b| = \sqrt{1^2 + 1^2} = \sqrt{2}$. $|c| = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}$. $\hat{u}_b = \frac{b}{|b|} = \frac{i+j}{\sqrt{2}}$. $\hat{u}_c = \frac{c}{|c|} = \frac{i-j+4k}{3\sqrt{2}}$.

3. Case 1: $a$ is parallel to $\hat{u}_b + \hat{u}_c$ (Internal Bisector)

   • $\hat{u}_b + \hat{u}_c = \frac{i+j}{\sqrt{2}} + \frac{i-j+4k}{3\sqrt{2}} = \frac{3(i+j) + (i-j+4k)}{3\sqrt{2}} = \frac{3i+3j+i-j+4k}{3\sqrt{2}} = \frac{4i+2j+4k}{3\sqrt{2}}$.
   • If $a = \lambda (\hat{u}_b + \hat{u}_c)$, then $\alpha i + 2j + \beta k = \lambda \left(\frac{4}{3\sqrt{2}}i + \frac{2}{3\sqrt{2}}j + \frac{4}{3\sqrt{2}}k\right)$.
   • Equating the coefficients of $j$: $2 = \lambda \frac{2}{3\sqrt{2}} \implies \lambda = 3\sqrt{2}$.
   • Equating coefficients of $i$ and $k$: $\alpha = (3\sqrt{2}) \frac{4}{3\sqrt{2}} = 4$. $\beta = (3\sqrt{2}) \frac{4}{3\sqrt{2}} = 4$.
   • Check Equation 1: $2\alpha - \beta = 2(4) - 4 = 8 - 4 = 4$. This solution is valid.
   • Test options with $a = 4i+2j+4k$: (a) $a \cdot k + 4 = (4i+2j+4k) \cdot k + 4 = 4+4=8 \neq 0$. (b) $a \cdot k + 2 = 4+2=6 \neq 0$. (c) $a \cdot i + 1 = (4i+2j+4k) \cdot i + 1 = 4+1=5 \neq 0$. (d) $a \cdot i + 3 = 4+3=7 \neq 0$.
   • This case does not yield a correct option.

4. Case 2: $a$ is parallel to $\hat{u}_b - \hat{u}_c$ (External Bisector)

   • $\hat{u}_b - \hat{u}_c = \frac{i+j}{\sqrt{2}} - \frac{i-j+4k}{3\sqrt{2}} = \frac{3(i+j) - (i-j+4k)}{3\sqrt{2}} = \frac{3i+3j-i+j-4k}{3\sqrt{2}} = \frac{2i+4j-4k}{3\sqrt{2}}$.
   • If $a = \lambda (\hat{u}_b - \hat{u}_c)$, then $\alpha i + 2j + \beta k = \lambda \left(\frac{2}{3\sqrt{2}}i + \frac{4}{3\sqrt{2}}j - \frac{4}{3\sqrt{2}}k\right)$.
   • Equating the coefficients of $j$: $2 = \lambda \frac{4}{3\sqrt{2}} \implies \lambda = \frac{3\sqrt{2}}{2}$.
   • Equating coefficients of $i$ and $k$: $\alpha = \left(\frac{3\sqrt{2}}{2}\right) \frac{2}{3\sqrt{2}} = 1$. $\beta = \left(\frac{3\sqrt{2}}{2}\right) \left(-\frac{4}{3\sqrt{2}}\right) = -2$.
   • Check Equation 1: $2\alpha - \beta = 2(1) - (-2) = 2+2=4$. This solution is valid.
   • Test options with $a = i+2j-2k$: (a) $a \cdot k + 4 = (i+2j-2k) \cdot k + 4 = -2+4=2 \neq 0$. (b) $a \cdot k + 2 = -2+2=0$. This option is correct. (c) $a \cdot i + 1 = (i+2j-2k) \cdot i + 1 = 1+1=2 \neq 0$. (d) $a \cdot i + 3 = 1+3=4 \neq 0$.

The correct option is (b).