Question:
\(\vec a,\vec b,\vec c\) are non-coplanar vectors. If
\[
\vec a+3\vec b+4\vec c
=
x(\vec a-2\vec b+3\vec c)
+y(\vec a+5\vec b-2\vec c)
+z(6\vec a+14\vec b+4\vec c),
\]
then
\[
x+y+z=
\]
\(\vec a,\vec b,\vec c\) are non-coplanar vectors. If
\[
\vec a+3\vec b+4\vec c
=
x(\vec a-2\vec b+3\vec c)
+y(\vec a+5\vec b-2\vec c)
+z(6\vec a+14\vec b+4\vec c),
\]
then
\[
x+y+z=
\]
Show Hint
When non-coplanar vectors are involved, treat them as linearly independent vectors. Equate the coefficients of \(\vec a,\vec b,\vec c\) separately to form simultaneous linear equations.
Updated On: Jun 22, 2026
- \(-5\)
- \(-4\)
- \(4\)
- \(5\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Use the fact that \(\vec a,\vec b,\vec c\) are non-coplanar.
Since \(\vec a,\vec b,\vec c\) are non-coplanar, they are linearly independent.
Therefore, coefficients of \(\vec a,\vec b,\vec c\) on both sides must be equal.
Step 2: Expand the right-hand side.
\[ x(\vec a-2\vec b+3\vec c) +y(\vec a+5\vec b-2\vec c) +z(6\vec a+14\vec b+4\vec c) \] \[ =(x+y+6z)\vec a+(-2x+5y+14z)\vec b+(3x-2y+4z)\vec c \]
Step 3: Compare coefficients.
From \[ \vec a+3\vec b+4\vec c = (x+y+6z)\vec a+(-2x+5y+14z)\vec b+(3x-2y+4z)\vec c, \] we get: \[ x+y+6z=1 \] \[ -2x+5y+14z=3 \] \[ 3x-2y+4z=4 \]
Step 4: Solve the system of equations.
From \[ x+y+6z=1, \] we get \[ x=1-y-6z \] Substitute this in the second equation: \[ -2(1-y-6z)+5y+14z=3 \] \[ -2+2y+12z+5y+14z=3 \] \[ 7y+26z=5 \] Substitute \(x=1-y-6z\) in the third equation: \[ 3(1-y-6z)-2y+4z=4 \] \[ 3-3y-18z-2y+4z=4 \] \[ -5y-14z=1 \]
Step 5: Find \(x+y+z\).
Now solve: \[ 7y+26z=5 \] and \[ -5y-14z=1 \] Multiplying the first equation by \(5\): \[ 35y+130z=25 \] Multiplying the second equation by \(7\): \[ -35y-98z=7 \] Adding, \[ 32z=32 \] \[ z=1 \] Then, \[ 7y+26=5 \] \[ 7y=-21 \] \[ y=-3 \] Now, \[ x=1-y-6z \] \[ x=1-(-3)-6(1) \] \[ x=-2 \] Therefore, \[ x+y+z=-2-3+1=-4 \]
Step 6: Final conclusion.
Hence, \[ \boxed{-4} \]
Since \(\vec a,\vec b,\vec c\) are non-coplanar, they are linearly independent.
Therefore, coefficients of \(\vec a,\vec b,\vec c\) on both sides must be equal.
Step 2: Expand the right-hand side.
\[ x(\vec a-2\vec b+3\vec c) +y(\vec a+5\vec b-2\vec c) +z(6\vec a+14\vec b+4\vec c) \] \[ =(x+y+6z)\vec a+(-2x+5y+14z)\vec b+(3x-2y+4z)\vec c \]
Step 3: Compare coefficients.
From \[ \vec a+3\vec b+4\vec c = (x+y+6z)\vec a+(-2x+5y+14z)\vec b+(3x-2y+4z)\vec c, \] we get: \[ x+y+6z=1 \] \[ -2x+5y+14z=3 \] \[ 3x-2y+4z=4 \]
Step 4: Solve the system of equations.
From \[ x+y+6z=1, \] we get \[ x=1-y-6z \] Substitute this in the second equation: \[ -2(1-y-6z)+5y+14z=3 \] \[ -2+2y+12z+5y+14z=3 \] \[ 7y+26z=5 \] Substitute \(x=1-y-6z\) in the third equation: \[ 3(1-y-6z)-2y+4z=4 \] \[ 3-3y-18z-2y+4z=4 \] \[ -5y-14z=1 \]
Step 5: Find \(x+y+z\).
Now solve: \[ 7y+26z=5 \] and \[ -5y-14z=1 \] Multiplying the first equation by \(5\): \[ 35y+130z=25 \] Multiplying the second equation by \(7\): \[ -35y-98z=7 \] Adding, \[ 32z=32 \] \[ z=1 \] Then, \[ 7y+26=5 \] \[ 7y=-21 \] \[ y=-3 \] Now, \[ x=1-y-6z \] \[ x=1-(-3)-6(1) \] \[ x=-2 \] Therefore, \[ x+y+z=-2-3+1=-4 \]
Step 6: Final conclusion.
Hence, \[ \boxed{-4} \]
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