Question 10

Mathematics Scalar and Vector Products Medium

If \(\vec{a}\) and \(\vec{b}\) are two vectors such that \(|\vec{a}|=3, |\vec{b}|=4\) and \(|\vec{a}+\vec{b}|=1\), then the value of \(|\vec{a}-\vec{b}|\) is

(A) 2
(B) 7
(C) 6
(D) 1
✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

We are given \(|\vec{a}|=3\), \(|\vec{b}|=4\), and \(|\vec{a}+\vec{b}|=1\). We want \(|\vec{a}-\vec{b}|\). Step 1: Use the identity \[ |\vec{a}+\vec{b}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2} + 2\vec{a}\cdot\vec{b}. \] Substitute values: \[ 1^{2} = 3^{2} + 4^{2} + 2\vec{a}\cdot\vec{b}, \] \[ 1 = 9 + 16 + 2\vec{a}\cdot\vec{b}, \] \[ 2\vec{a}\cdot\vec{b} = 1 - 25 = -24, \] \[ \vec{a}\cdot\vec{b} = -12. \] Step 2: Use the formula for difference \[ |\vec{a}-\vec{b}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2} - 2\vec{a}\cdot\vec{b}. \] Substitute: \[ |\vec{a}-\vec{b}|^{2} = 9 + 16 - 2(-12), \] \[ = 25 + 24 = 49. \] Final Answer: \[ |\vec{a}-\vec{b}| = \sqrt{49} = 7. \] \[ \boxed{7} \]

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