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: Understanding the Question:
We are told that vector $\vec{r}$ can be expressed as a linear combination of $\vec{a}$ and $\vec{b}$. This means $\vec{r} = x\vec{a} + y\vec{b}$ for some scalars $x$ and $y$. We need to find the values of $x$ and $y$.

Step 2: Key Formula or Approach:
Set up the vector equation $\vec{r} = x\vec{a} + y\vec{b}$. Equate the respective $\hat{i}$, $\hat{j}$, and $\hat{k}$ components to form a system of linear equations and solve for $x$ and $y$.

Step 3: Detailed Explanation:
Write out the full vector equation:
$$-4\hat{i} - 6\hat{j} - 2\hat{k} = x(-\hat{i} + 4\hat{j} + 3\hat{k}) + y(-8\hat{i} - \hat{j} + 3\hat{k})$$ Group the components on the right side:
$$-4\hat{i} - 6\hat{j} - 2\hat{k} = (-x - 8y)\hat{i} + (4x - y)\hat{j} + (3x + 3y)\hat{k}$$ Equating corresponding components yields three equations:
1. $-x - 8y = -4$
2. $4x - y = -6$
3. $3x + 3y = -2$
From equation (2), solve for $y$:
$$y = 4x + 6$$ Substitute this expression for $y$ into equation (1):
$$-x - 8(4x + 6) = -4$$ $$-x - 32x - 48 = -4$$ $$-33x = 44$$ $$x = -\frac{44}{33} = -\frac{4}{3}$$ Now substitute $x = -\frac{4}{3}$ back into the expression for $y$:
$$y = 4\left(-\frac{4}{3}\right) + 6 = -\frac{16}{3} + \frac{18}{3} = \frac{2}{3}$$ Verify with equation (3):
$$3\left(-\frac{4}{3}\right) + 3\left(\frac{2}{3}\right) = -4 + 2 = -2$$ (This matches perfectly).
Thus, the linear combination is $\vec{r} = -\frac{4}{3}\vec{a} + \frac{2}{3}\vec{b}$.

Step 4: Final Answer:
The relation is $\vec{r} = -\frac{4}{3}\vec{a} + \frac{2}{3}\vec{b}$, matching option (A).