Question

If $\vec{a}=2\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}-\hat{j}+\hat{k}$, then the value of $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})$ is equal to

• A. 8
• B. 7
• C. 9
• D. 11
• E. 13

Hint:

Vector Tip: Treat $(\vec{x}+\vec{y})\cdot(\vec{x}-\vec{y})$ exactly like $(x+y)(x-y) = x^2 - y^2$ in standard algebra! It skips the tedious process of calculating new combined vectors.

Answer 1

The Correct Option is D

Concept:
The dot product operation is distributive over vector addition. The expression $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})$ expands just like a difference of squares in algebra: $\vec{a}\cdot\vec{a} - \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{a} - \vec{b}\cdot\vec{b}$. Since the dot product is commutative, this simplifies perfectly to $|\vec{a}|^2 - |\vec{b}|^2$.

Step 1: Apply the difference of squares identity.
Instead of adding and subtracting the vectors first, use the property: $$(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2 - |\vec{b}|^2$$

Step 2: Calculate the squared magnitude of vector a.
Using $\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}$: $$|\vec{a}|^2 = (2)^2 + (2)^2 + (3)^2$$ $$|\vec{a}|^2 = 4 + 4 + 9 = 17$$

Step 3: Calculate the squared magnitude of vector b.
Using $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$: $$|\vec{b}|^2 = (2)^2 + (-1)^2 + (1)^2$$ $$|\vec{b}|^2 = 4 + 1 + 1 = 6$$

Step 4: Substitute values into the simplified expression.
Plug the squared magnitudes back into our identity from
Step 1: $$(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = 17 - 6$$

Step 5: Perform the final subtraction.
Calculate the difference to find the final value: $$17 - 6 = 11$$ Hence the correct answer is (D) 11.