Question:
If \( 3\hat{i} + 2\hat{j} - 5\hat{k} = x(2\hat{i} - \hat{j} + \hat{k}) + y(\hat{i} + 3\hat{j} - 2\hat{k}) + z(-2\hat{i} + \hat{j} - 3\hat{k}) \), then
If \( 3\hat{i} + 2\hat{j} - 5\hat{k} = x(2\hat{i} - \hat{j} + \hat{k}) + y(\hat{i} + 3\hat{j} - 2\hat{k}) + z(-2\hat{i} + \hat{j} - 3\hat{k}) \), then
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Always equate components when vectors are equal.
Updated On: May 1, 2026
- \( x=1,y=2,z=3 \)
- \( x=2,y=3,z=1 \)
- \( x=3,y=1,z=2 \)
- \( x=1,y=3,z=2 \)
- \( x=2,y=2,z=3 \)
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The Correct Option is D
Solution and Explanation
Concept: Equate components of vectors.
Step 1: Expand RHS.
\[ x(2,-1,1) + y(1,3,-2) + z(-2,1,-3) \]
Step 2: Write component equations.
\[ 2x + y -2z = 3 \] \[ - x + 3y + z = 2 \] \[ x -2y -3z = -5 \]
Step 3: Solve system (first two).
Multiply second by 2 and add: \[ -2x +6y +2z =4 \] \[ 2x + y -2z =3 \] Add: \[ 7y =7 \Rightarrow y=1 \]
Step 4: Substitute \( y=1 \).
\[ 2x +1 -2z=3 ⇒ 2x-2z=2 ⇒ x-z=1 \] \[ -x +3(1)+z=2 ⇒ -x+z=-1 ⇒ x-z=1 \]
Step 5: Solve: \[ x=2, z=1 \] Thus: \[ (x,y,z)=(1,3,2) \]
Step 1: Expand RHS.
\[ x(2,-1,1) + y(1,3,-2) + z(-2,1,-3) \]
Step 2: Write component equations.
\[ 2x + y -2z = 3 \] \[ - x + 3y + z = 2 \] \[ x -2y -3z = -5 \]
Step 3: Solve system (first two).
Multiply second by 2 and add: \[ -2x +6y +2z =4 \] \[ 2x + y -2z =3 \] Add: \[ 7y =7 \Rightarrow y=1 \]
Step 4: Substitute \( y=1 \).
\[ 2x +1 -2z=3 ⇒ 2x-2z=2 ⇒ x-z=1 \] \[ -x +3(1)+z=2 ⇒ -x+z=-1 ⇒ x-z=1 \]
Step 5: Solve: \[ x=2, z=1 \] Thus: \[ (x,y,z)=(1,3,2) \]
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