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 \((x,y,z)=\)

• A. \((1,-3,2)\)
• B. \((-1,3,-2)\)
• C. \((1,3,2)\)
• D. \((1,3,-2)\)

Hint:

Dot product conditions involving an unknown vector usually lead to a system of linear equations in its components. Solve the equations simultaneously.

Answer 1

The Correct Option is D

Step 1: Write out the three dot product equations.
Let $ \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} $. The three conditions give: \[ 2x + 3y + z = 9 \quad (1) \] \[ 4x + y = 7 \quad (2) \] \[ x - 3y - 7z = 6 \quad (3) \]
Step 2: Express $ y $ in terms of $ x $ from equation (2).
From (2): $ y = 7 - 4x $. This gives us $ y $ once we know $ x $, reducing the system to two unknowns.
Step 3: Substitute into equation (1) to find $ z $.
Substitute $ y = 7 - 4x $ into (1): \[ 2x + 3(7 - 4x) + z = 9 \Rightarrow 2x + 21 - 12x + z = 9 \Rightarrow z = 10x - 12 \]
Step 4: Substitute both into equation (3) to find $ x $.
\[ x - 3(7-4x) - 7(10x-12) = 6 \] \[ x - 21 + 12x - 70x + 84 = 6 \] \[ -57x + 63 = 6 \Rightarrow -57x = -57 \Rightarrow x = 1 \]
Step 5: Back-substitute to find $ y $ and $ z $.
$ y = 7 - 4(1) = 3 $ and $ z = 10(1) - 12 = -2 $. Verify in all three original equations - all are satisfied.
Step 6: State the answer.
\[ \boxed{(x, y, z) = (1, 3, -2)} \]