Question

A unit vector \( \mathbf{e} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is coplanar with the vectors \( \mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \) and \( 3\mathbf{i} + \mathbf{j} - 5\mathbf{k} \) and is perpendicular to the vector \( \mathbf{i} + \mathbf{j} + \mathbf{k} \). Calculate \(2a^2 + 3b^2 + 4c^2\):

• A. 1
• B. 3
• C. -1
• D. \(\sqrt{2}\) \vspace{0.5cm}

Hint:

When solving vector problems involving conditions like perpendicularity and coplanarity, use dot products and scalar triple products efficiently. Be careful with algebraic manipulations to solve the system of equations.

Answer 1

The Correct Option is B

Step 1: Use the condition that \( \mathbf{e} \) is perpendicular to \( \mathbf{i} + \mathbf{j} + \mathbf{k} \). Since \( \mathbf{e} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is perpendicular to the vector \( \mathbf{i} + \mathbf{j} + \mathbf{k} \), we have the dot product condition: \[ \mathbf{e} \cdot (\mathbf{i} + \mathbf{j} + \mathbf{k}) = 0 \] \[ a + b + c = 0 \] Step 2: Use the condition that \( \mathbf{e} \) is coplanar with \( \mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \) and \( 3\mathbf{i} + \mathbf{j} - 5\mathbf{k} \). For coplanarity, the scalar triple product of \( \mathbf{e}, \mathbf{i} - 3\mathbf{j} + 5\mathbf{k}, \) and \( 3\mathbf{i} + \mathbf{j} - 5\mathbf{k} \) must be zero: \[ \begin{vmatrix} a & b & c
1 & -3 & 5
3 & 1 & -5 \end{vmatrix} = 0 \] Expanding the determinant: \[ a \left( (-3)(-5) - (5)(1) \right) - b \left( (1)(-5) - (5)(3) \right) + c \left( (1)(1) - (-3)(3) \right) = 0 \] \[ a(15 - 5) - b(-5 - 15) + c(1 + 9) = 0 \] \[ a(10) - b(-20) + c(10) = 0 \] \[ 10a + 20b + 10c = 0 \] Step 3: Solve the system of equations. We now have the system of equations: 1. \( a + b + c = 0 \) 2. \( 10a + 20b + 10c = 0 \) From the first equation, solve for \( c \): \[ c = -a - b \] Substitute into the second equation: \[ 10a + 20b + 10(-a - b) = 0 \] \[ 10a + 20b - 10a - 10b = 0 \] \[ 10b = 0 \] \[ b = 0 \] Substitute \( b = 0 \) into \( a + b + c = 0 \): \[ a + c = 0 \quad \Rightarrow \quad c = -a \] Step 4: Use the unit vector condition. Since \( \mathbf{e} \) is a unit vector: \[ a^2 + b^2 + c^2 = 1 \] Substitute \( b = 0 \) and \( c = -a \): \[ a^2 + 0^2 + (-a)^2 = 1 \] \[ 2a^2 = 1 \quad \Rightarrow \quad a^2 = \frac{1}{2} \] \[ a = \pm \frac{1}{\sqrt{2}} \] Now, using \( c = -a \), we find: \[ c = \mp \frac{1}{\sqrt{2}} \] Step 5: Compute \( 2a^2 + 3b^2 + 4c^2 \). Substitute \( a^2 = \frac{1}{2} \), \( b = 0 \), and \( c^2 = \frac{1}{2} \) into the expression \( 2a^2 + 3b^2 + 4c^2 \): \[ 2a^2 + 3b^2 + 4c^2 = 2 \times \frac{1}{2} + 3 \times 0 + 4 \times \frac{1}{2} \] \[ = 1 + 0 + 2 = 3 \] Thus, the correct value of \( 2a^2 + 3b^2 + 4c^2 \) is \( \boxed{3} \). \vspace{0.5cm}