Question

Let $\vec{a} = \hat{i} +2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}-3\hat{j}+\hat{k}$ and $\vec{c}=3\hat{i}+\hat{j}-2\hat{k}$ be three vectors. If $\vec{r}$ is a vector such that $\vec{r}\cdot\vec{a} = 0$, $\vec{r}\cdot\vec{b} = -2$ and $\vec{r}\cdot\vec{c} = 6$ then $\vec{r}\cdot(3\hat{i}+\hat{j}+\hat{k})= $

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

When a vector $\vec{r}$ is defined by its dot products with three other non-coplanar vectors, it creates a system of three linear equations. Solving this system gives the components of $\vec{r}$.

Answer 1

The Correct Option is D

Let the vector $\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}$.
We are given three conditions based on the dot product:
1) $\vec{r} \cdot \vec{a} = (x\hat{i}+y\hat{j}+z\hat{k}) \cdot (\hat{i}+2\hat{j}+3\hat{k}) = x+2y+3z=0$. (Eq. 1)
2) $\vec{r} \cdot \vec{b} = (x\hat{i}+y\hat{j}+z\hat{k}) \cdot (2\hat{i}-3\hat{j}+\hat{k}) = 2x-3y+z=-2$. (Eq. 2)
3) $\vec{r} \cdot \vec{c} = (x\hat{i}+y\hat{j}+z\hat{k}) \cdot (3\hat{i}+\hat{j}-2\hat{k}) = 3x+y-2z=6$. (Eq. 3)
We need to solve this system of linear equations for $x, y, z$.
From Eq. 1, $x = -2y-3z$. Substitute this into Eq. 2 and Eq. 3.
Into Eq. 2: $2(-2y-3z) - 3y + z = -2 \implies -4y-6z-3y+z=-2 \implies -7y-5z=-2 \implies 7y+5z=2$. (Eq. 4)
Into Eq. 3: $3(-2y-3z) + y - 2z = 6 \implies -6y-9z+y-2z=6 \implies -5y-11z=6 \implies 5y+11z=-6$. (Eq. 5)
Now we solve the system for $y$ and $z$ from Eq. 4 and Eq. 5.
Multiply Eq. 4 by 5 and Eq. 5 by 7:
$35y + 25z = 10$
$35y + 77z = -42$
Subtracting the second new equation from the first: $(35y+25z) - (35y+77z) = 10 - (-42) \implies -52z = 52 \implies z=-1$.
Substitute $z=-1$ into Eq. 4: $7y+5(-1)=2 \implies 7y-5=2 \implies 7y=7 \implies y=1$.
Substitute $y=1$ and $z=-1$ into the expression for $x$: $x = -2(1)-3(-1) = -2+3=1$.
So, the vector is $\vec{r} = 1\hat{i}+1\hat{j}-1\hat{k}$.
Finally, we need to calculate $\vec{r}\cdot(3\hat{i}+\hat{j}+\hat{k})$.
$(\hat{i}+\hat{j}-\hat{k}) \cdot (3\hat{i}+\hat{j}+\hat{k}) = (1)(3) + (1)(1) + (-1)(1) = 3+1-1=3$.