Question

If $\vec{r} = -4\hat{i} - 6\hat{j} - 2\hat{k}$ is a linear combination of the vectors $\vec{a} = -\hat{i} + 4\hat{j} + 3\hat{k}$ and $\vec{b} = -8\hat{i} - \hat{j} + 3\hat{k}$, then

• A. $\vec{r} = -\frac{4}{3}\vec{a} + \frac{2}{3}\vec{b}$
• B. $\vec{r} = \frac{4}{3}\vec{a} + \frac{2}{3}\vec{b}$
• C. $\vec{r} = -\frac{1}{3}\vec{a} + \frac{2}{3}\vec{b}$
• D. $\vec{r} = \frac{1}{3}\vec{a} - \frac{1}{3}\vec{b}$

Hint:

To save time during an exam, instead of fully solving the system of equations, you can quickly plug the scalar coefficients from the options into the original vectors to see which one correctly yields $\vec{r}$'s $\hat{k}$ component.

Answer 1

The Correct Option is A

Step 1: Understand "linear combination".
Saying $\vec r$ is a linear combination of $\vec a$ and $\vec b$ means $\vec r=x\,\vec a+y\,\vec b$ for some numbers $x$ and $y$. We must find these numbers.

Step 2: Write the vector equation.
\[ -4\hat i-6\hat j-2\hat k=x(-\hat i+4\hat j+3\hat k)+y(-8\hat i-\hat j+3\hat k). \]

Step 3: Group by direction.
\[ -4\hat i-6\hat j-2\hat k=(-x-8y)\hat i+(4x-y)\hat j+(3x+3y)\hat k. \]

Step 4: Compare each direction.
Matching $\hat i$, $\hat j$, $\hat k$ gives three equations:
(1) $-x-8y=-4$, (2) $4x-y=-6$, (3) $3x+3y=-2$.

Step 5: Solve two of them.
From (2): $y=4x+6$. Put this into (1): \[ -x-8(4x+6)=-4\Rightarrow -x-32x-48=-4\Rightarrow -33x=44, \] so $x=-\frac{4}{3}$.

Step 6: Find $y$.
\[ y=4\!\left(-\tfrac{4}{3}\right)+6=-\tfrac{16}{3}+\tfrac{18}{3}=\tfrac{2}{3}. \]

Step 7: Check with the third equation.
$3\!\left(-\frac43\right)+3\!\left(\frac23\right)=-4+2=-2$. It matches, so the answer is reliable. Hence $\vec r=-\frac{4}{3}\vec a+\frac{2}{3}\vec b$, option (1).
\[ \boxed{\vec r=-\tfrac{4}{3}\vec a+\tfrac{2}{3}\vec b} \]