Question:
Let
\[
\vec{a}=2\hat{i}+3\hat{j}+\hat{k},\quad
\vec{b}=4\hat{i}+\hat{j},\quad
\vec{c}=\hat{i}-3\hat{j}-7\hat{k}
\]
If
\[
\vec{r}=x\hat{i}+y\hat{j}+z\hat{k},
\]
\[
\vec{r}\cdot\vec{a}=9,\quad
\vec{r}\cdot\vec{b}=7,\quad
\vec{r}\cdot\vec{c}=6,
\]
then \((x,y,z)=\)
Let
\[
\vec{a}=2\hat{i}+3\hat{j}+\hat{k},\quad
\vec{b}=4\hat{i}+\hat{j},\quad
\vec{c}=\hat{i}-3\hat{j}-7\hat{k}
\]
If
\[
\vec{r}=x\hat{i}+y\hat{j}+z\hat{k},
\]
\[
\vec{r}\cdot\vec{a}=9,\quad
\vec{r}\cdot\vec{b}=7,\quad
\vec{r}\cdot\vec{c}=6,
\]
then \((x,y,z)=\)
Show Hint
Dot product conditions involving an unknown vector usually lead to a system of linear equations in its components. Solve the equations simultaneously.
Updated On: Jun 25, 2026
- \((1,-3,2)\)
- \((-1,3,-2)\)
- \((1,3,2)\)
- \((1,3,-2)\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use the condition \(\vec{r}\cdot\vec{a}=9\).
\[ (x\hat{i}+y\hat{j}+z\hat{k})\cdot(2\hat{i}+3\hat{j}+\hat{k})=9 \] \[ 2x+3y+z=9 \]
Step 2: Use the condition \(\vec{r}\cdot\vec{b}=7\).
\[ (x\hat{i}+y\hat{j}+z\hat{k})\cdot(4\hat{i}+\hat{j})=7 \] \[ 4x+y=7 \]
Step 3: Use the condition \(\vec{r}\cdot\vec{c}=6\).
\[ (x\hat{i}+y\hat{j}+z\hat{k})\cdot(\hat{i}-3\hat{j}-7\hat{k})=6 \] \[ x-3y-7z=6 \] Thus, we obtain \[ 2x+3y+z=9 \] \[ 4x+y=7 \] \[ x-3y-7z=6 \]
Step 4: Solve the system of equations.
From \[ 4x+y=7 \] \[ y=7-4x \] Substituting in \[ 2x+3y+z=9 \] \[ 2x+3(7-4x)+z=9 \] \[ z=10x-12 \] Now substitute \(y=7-4x\) and \(z=10x-12\) into \[ x-3y-7z=6 \] \[ x-3(7-4x)-7(10x-12)=6 \] \[ x-21+12x-70x+84=6 \] \[ -57x+63=6 \] \[ x=1 \] Hence, \[ y=7-4(1)=3 \] and \[ z=10(1)-12=-2 \] Therefore, \[ (x,y,z)=(1,3,-2) \]
Step 5: Final conclusion.
\[ \boxed{(x,y,z)=(1,3,-2)} \]
\[ (x\hat{i}+y\hat{j}+z\hat{k})\cdot(2\hat{i}+3\hat{j}+\hat{k})=9 \] \[ 2x+3y+z=9 \]
Step 2: Use the condition \(\vec{r}\cdot\vec{b}=7\).
\[ (x\hat{i}+y\hat{j}+z\hat{k})\cdot(4\hat{i}+\hat{j})=7 \] \[ 4x+y=7 \]
Step 3: Use the condition \(\vec{r}\cdot\vec{c}=6\).
\[ (x\hat{i}+y\hat{j}+z\hat{k})\cdot(\hat{i}-3\hat{j}-7\hat{k})=6 \] \[ x-3y-7z=6 \] Thus, we obtain \[ 2x+3y+z=9 \] \[ 4x+y=7 \] \[ x-3y-7z=6 \]
Step 4: Solve the system of equations.
From \[ 4x+y=7 \] \[ y=7-4x \] Substituting in \[ 2x+3y+z=9 \] \[ 2x+3(7-4x)+z=9 \] \[ z=10x-12 \] Now substitute \(y=7-4x\) and \(z=10x-12\) into \[ x-3y-7z=6 \] \[ x-3(7-4x)-7(10x-12)=6 \] \[ x-21+12x-70x+84=6 \] \[ -57x+63=6 \] \[ x=1 \] Hence, \[ y=7-4(1)=3 \] and \[ z=10(1)-12=-2 \] Therefore, \[ (x,y,z)=(1,3,-2) \]
Step 5: Final conclusion.
\[ \boxed{(x,y,z)=(1,3,-2)} \]
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