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 \(2a^2 + 3b^2 + 4c^2\):

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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.
Updated On: Mar 17, 2025
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The Correct Option is B

Solution and Explanation

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}
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