Question

Let $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$. Then $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) =$

• A. 6
• B. 7
• C. 8
• D. 9
• E. 10

Hint:

The difference of squares identity works for vectors in a dot product just like it does for real numbers.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are asked to compute the dot product of the sum and difference of two vectors. We can approach this in two ways: either by first calculating the sum and difference vectors and then taking their dot product, or by using the algebraic properties of the dot product.
Step 2: Key Formula or Approach:
Method 1: Direct Calculation
1. Compute \( \vec{a} + \vec{b} \).
2. Compute \( \vec{a} - \vec{b} \).
3. Compute the dot product of the two resulting vectors.
Method 2: Algebraic Simplification
Use the distributive property of the dot product, which works like the algebraic identity \( (x+y)(x-y) = x^2 - y^2 \).
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} \] Since \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \) (dot product is commutative), this simplifies to:
\[ \vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{b} = |\vec{a}|^2 - |\vec{b}|^2 \] Step 3: Detailed Explanation:
We will use Method 2 as it is generally faster.
Given vectors:
\( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \)
\( \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \)
First, calculate the square of the magnitude of \( \vec{a} \):
\[ |\vec{a}|^2 = (3)^2 + (2)^2 + (2)^2 = 9 + 4 + 4 = 17 \] Next, calculate the square of the magnitude of \( \vec{b} \):
\[ |\vec{b}|^2 = (1)^2 + (2)^2 + (-2)^2 = 1 + 4 + 4 = 9 \] Now, use the formula \( |\vec{a}|^2 - |\vec{b}|^2 \):
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 17 - 9 = 8 \] Verification with Method 1:
\( \vec{a} + \vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j} + 0\hat{k} \)
\( \vec{a} - \vec{b} = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} = 2\hat{i} + 0\hat{j} + 4\hat{k} \)
Now, take the dot product:
\[ (4\hat{i} + 4\hat{j}) \cdot (2\hat{i} + 4\hat{k}) = (4)(2) + (4)(0) + (0)(4) = 8 + 0 + 0 = 8 \] Both methods yield the same result.
Step 4: Final Answer:
The value of \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) \) is 8.