Question

If $|\vec{a}|=2$, $|\vec{b}|=3$ and $\vec{a}\cdot\vec{b}=4$, then $|\vec{a}-\vec{b}|$ is equal to

• A. $\sqrt{5}$
• B. $\sqrt{7}$
• C. $\sqrt{6}$
• D. 5
• E. 6

Hint:

Formula Tip: The expansion $|\vec{a} \pm \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 \pm 2(\vec{a}\cdot\vec{b})$ is exactly analogous to the Law of Cosines $c^2 = a^2 + b^2 - 2ab\cos\theta$ in trigonometry!

Answer 1

The Correct Option is A

Concept:
To find the magnitude of the difference of two vectors, we utilize the property that the square of a vector's magnitude is equal to the dot product of the vector with itself: $|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$. This expands into an algebraic formula involving magnitudes and the dot product.

Step 1: State the expansion formula.
Expand the squared magnitude expression: $$|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$$

Step 2: Identify the given values.
From the problem statement, we are provided with: $$|\vec{a}| = 2$$ $$|\vec{b}| = 3$$ $$\vec{a} \cdot \vec{b} = 4$$

Step 3: Substitute values into the formula.
Plug the known variables into the expansion from
Step 1: $$|\vec{a} - \vec{b}|^2 = (2)^2 + (3)^2 - 2(4)$$

Step 4: Calculate the squared magnitude.
Perform the arithmetic operations: $$|\vec{a} - \vec{b}|^2 = 4 + 9 - 8$$ $$|\vec{a} - \vec{b}|^2 = 13 - 8 = 5$$

Step 5: Find the final magnitude.
Take the square root of both sides to remove the square and find the actual magnitude: $$|\vec{a} - \vec{b}| = \sqrt{5}$$ Hence the correct answer is (A) $\sqrt{5$}.