Question

Given that $ \vec{a}, \vec{b}, \vec{c} $ are non-coplanar vectors and: $$ \vec{a} + 3\vec{b} + 4\vec{c} = x(\vec{a} - 2\vec{b} + 3\vec{c}) + y(\vec{a} + 5\vec{b} - 2\vec{c}) + z(6\vec{a} + 14\vec{b} + 4\vec{c}) $$ Find $ x + y + z $

• A. -5
• B. -4
• C. 4
• D. 5

Hint:

Equating vector coefficients gives a solvable linear system when vectors are linearly independent (non-coplanar).

Answer 1

The Correct Option is B

We are given an identity involving non-coplanar vectors. Hence, we equate coefficients of corresponding vectors on both sides. LHS: \[ \vec{a} + 3\vec{b} + 4\vec{c} \] RHS: Expand: \[ x(\vec{a} - 2\vec{b} + 3\vec{c}) + y(\vec{a} + 5\vec{b} - 2\vec{c}) + z(6\vec{a} + 14\vec{b} + 4\vec{c}) \] Break down:
- \( \vec{a} \): \( x + y + 6z \)
- \( \vec{b} \): \( -2x + 5y + 14z \)
- \( \vec{c} \): \( 3x - 2y + 4z \) Now equate coefficients:
1. Coefficient of \( \vec{a} \): \[ x + y + 6z = 1 \quad \text{(Equation 1)} \] 2. Coefficient of \( \vec{b} \): \[ -2x + 5y + 14z = 3 \quad \text{(Equation 2)} \] 3. Coefficient of \( \vec{c} \): \[ 3x - 2y + 4z = 4 \quad \text{(Equation 3)} \] Solve this system using substitution or matrix method. From (1): \( x = 1 - y - 6z \) Substitute into (2): \[ -2(1 - y - 6z) + 5y + 14z = 3 \Rightarrow -2 + 2y + 12z + 5y + 14z = 3 \Rightarrow 7y + 26z = 5 \quad \text{(Equation A)} \] Substitute into (3): \[ 3(1 - y - 6z) - 2y + 4z = 4 \Rightarrow 3 - 3y - 18z - 2y + 4z = 4 \Rightarrow -5y -14z = 1 \quad \text{(Equation B)} \] Now solve A and B: From A: \( 7y = 5 - 26z \Rightarrow y = \frac{5 - 26z}{7} \) Sub into B: \[ -5 \cdot \frac{5 - 26z}{7} -14z = 1 \Rightarrow -\frac{25 - 130z}{7} -14z = 1 \Rightarrow -\frac{25}{7} + \frac{130z}{7} -14z = 1 \Rightarrow \frac{130z - 98z}{7} = 1 + \frac{25}{7} = \frac{32}{7} \Rightarrow 32z = 32 \Rightarrow z = 1 \] Then, \[ y = \frac{5 - 26(1)}{7} = \frac{-21}{7} = -3,\quad x = 1 - y - 6z = 1 + 3 - 6 = -2 \] So: \[ x + y + z = -2 -3 + 1 = \boxed{-4} \]