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NDA 2013
Exam
NDA
Year
2013
Subject
Mathematics
Chapter
Vector Algebra

NDA 2013 · Mathematics · Vector Algebra

If the angle between the vectors i -m j and j + k is (π)/(3) , then...

If the angle between the vectors i^−mj^\hat{i}-m \hat{j} and j^+k^\hat{j}+\hat{k} is π3\frac{\pi}{3}, then what is the value of mm ?

Options

  1. A0
  2. B2
  3. C–2
  4. DNone of the above

Correct answer

D. None of the above

Step-by-step solution

Let,A→=i^−mj^ and  B→=j^+k^A→⋅B→=∣A→∣⋅∣B→∣⋅cos⁡(θ)\begin{aligned} {Let,} \overrightarrow{\mathrm{A}} & =\hat{i}-m \hat{j} \quad \text{\text{ and } } \quad \overrightarrow{\mathrm{B}}=\hat{j}+\hat{k} \\ \overrightarrow{\mathrm{ A}} \cdot \overrightarrow{\mathrm{ B}} & =|\overrightarrow{\mathrm{A}}| \cdot|\overrightarrow{\mathrm{B}}| \cdot \cos (\theta) \end{aligned}(i^−mj^)⋅(j^+k^)=∣i^−mj^∣⋅∣j^+k^∣⋅cos⁡(π3)−m=∣1+m2∣⋅∣12+12∣⋅12 or, −2m=1+m2\begin{aligned} (\hat{i}-m \hat{j}) \cdot(\hat{j}+\hat{k}) & =|\hat{i}-m \hat{j}| \cdot|\hat{j}+\hat{k}| \cdot \cos \left(\frac{\pi}{3}\right) \\ -m & =\left|\sqrt{1+m^2}\right| \cdot\left|\sqrt{1^2+1^2}\right| \cdot \frac{1}{2} \\ \text{ or, } -\sqrt{2} m & =\sqrt{1+m^2} \end{aligned}Squaringbothsides,wegetSquaring both sides, we get2m2=1+m2⇒m2=1m=±1\begin{aligned} 2 m^2 & =1+m^2 \Rightarrow m^2=1 \\ m & = \pm 1 \end{aligned}$
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