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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Vector Algebra

NDA 2015 · Mathematics · Vector Algebra

If | a |=7,| b |=11 and | a + b |=10 3 , then | a - b | is equal to

If ∣a⃗∣=7,∣b⃗∣=11|\vec{a}|=7,|\vec{b}|=11 and ∣a⃗+b⃗∣=103|\vec{a}+\vec{b}|=10 \sqrt{3}, then ∣a⃗−b⃗∣|\vec{a}-\vec{b}| is equal to

Options

  1. A40
  2. B10
  3. C4104 \sqrt{10}
  4. D2102 \sqrt{10}

Correct answer

D. 2102 \sqrt{10}

Step-by-step solution

|\overrightarrow{\mathrm{a}}|=7,|\overrightarrow{\mathrm{ b}}|=11 \text{ and }|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=10 \sqrt{3} Now ∣a⃗+b⃗∣2=∣a⃗∣2+∣b⃗∣2+2∣a⃗∣∣b⃗∣cos⁡θ|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}||\vec{b}| \cos \theta∴(103)2=49+121+2×7×7cos⁡θ∴300=170+154cos⁡θ300−170154=cos⁡θ∴6577=cos⁡θ\begin{array}{ll}\therefore & (10 \sqrt{3})^2=49+121+2 \times 7 \times 7 \cos \theta \\ \therefore & 300=170+154 \cos \theta \\ & \frac{300-170}{154}=\cos \theta \\ \therefore & \frac{65}{77}=\cos \theta\end{array}$\begin{aligned} & \text{ Now, }|\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}| \cos \theta \\ & =(7)^2+(11)^2-2 \times 7 \times 11 \times \frac{65}{77} \\ & =49+121-2 \times 65 \\ & =170-130=40 \\ \therefore & |\vec{a}-\vec{b}|=\sqrt{40}=2 \sqrt{10} \end{aligned}\therefore \quad$ Option (D) is correct.
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