Question

Let $|\bar{A}_1| = 3, |\bar{A}_2| = 5$ and $|\bar{A}_1 + \bar{A}_2| = 5$. The value of $(2\bar{A}_1 + 3\bar{A}_2).(3\bar{A}_1 - 2\bar{A}_2)$ is :-

• A. -112.5
• B. -106.5
• C. -118.5
• D. -99.5

Answer 1

Answer: (C)

-118.5

Answer 2

To solve the problem, we first need to find the dot product of $\bar{A}_1$ and $\bar{A}_2$ using the given magnitudes.

Given: $|\bar{A}_1| = 3$ $|\bar{A}_2| = 5$ $|\bar{A}_1 + \bar{A}_2| = 5$

We know the formula for the magnitude of the sum of two vectors: $|\bar{A}_1 + \bar{A}_2|^2 = |\bar{A}_1|^2 + |\bar{A}_2|^2 + 2\bar{A}_1 \cdot \bar{A}_2$

Substitute the given values into the formula: $5^2 = 3^2 + 5^2 + 2\bar{A}_1 \cdot \bar{A}_2$ $25 = 9 + 25 + 2\bar{A}_1 \cdot \bar{A}_2$ $25 = 34 + 2\bar{A}_1 \cdot \bar{A}_2$

Now, solve for $2\bar{A}_1 \cdot \bar{A}_2$: $2\bar{A}_1 \cdot \bar{A}_2 = 25 - 34$ $2\bar{A}_1 \cdot \bar{A}_2 = -9$ $\bar{A}_1 \cdot \bar{A}_2 = -\frac{9}{2} = -4.5$

Next, we need to evaluate the expression $(2\bar{A}_1 + 3\bar{A}_2).(3\bar{A}_1 - 2\bar{A}_2)$. We can expand this dot product using the distributive property, similar to algebraic multiplication: $(2\bar{A}_1 + 3\bar{A}_2).(3\bar{A}_1 - 2\bar{A}_2) = (2\bar{A}_1) \cdot (3\bar{A}_1) + (2\bar{A}_1) \cdot (-2\bar{A}_2) + (3\bar{A}_2) \cdot (3\bar{A}_1) + (3\bar{A}_2) \cdot (-2\bar{A}_2)$ $= 6(\bar{A}_1 \cdot \bar{A}_1) - 4(\bar{A}_1 \cdot \bar{A}_2) + 9(\bar{A}_2 \cdot \bar{A}_1) - 6(\bar{A}_2 \cdot \bar{A}_2)$

Recall that $\bar{A} \cdot \bar{A} = |\bar{A}|^2$ and $\bar{A}_1 \cdot \bar{A}_2 = \bar{A}_2 \cdot \bar{A}_1$. So, the expression becomes: $= 6|\bar{A}_1|^2 - 4(\bar{A}_1 \cdot \bar{A}_2) + 9(\bar{A}_1 \cdot \bar{A}_2) - 6|\bar{A}_2|^2$ Combine the terms with $\bar{A}_1 \cdot \bar{A}_2$: $= 6|\bar{A}_1|^2 + 5(\bar{A}_1 \cdot \bar{A}_2) - 6|\bar{A}_2|^2$

Now, substitute the known values: $|\bar{A}_1| = 3$, $|\bar{A}_2| = 5$, and $\bar{A}_1 \cdot \bar{A}_2 = -4.5$: $= 6(3^2) + 5(-4.5) - 6(5^2)$ $= 6(9) + (-22.5) - 6(25)$ $= 54 - 22.5 - 150$ $= 54 - 172.5$ $= -118.5$