Three Dimensional Geometry — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
Find the angle between the vectors whose direction cosines are proportional to 2, 3, -6 and 3, -4, 5.
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Answer
Let $\vec{\text{a}}$ be a vector with direction ratios 2, 3, -6.
$\Rightarrow\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
Let $\vec{\text{b}}$ be a vector with direction ratios 3, -4, 5.
$\Rightarrow\vec{\text{b}}=3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$
Let $\theta$ be the angle between the given vectors.
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}).(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})}{\big|2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}\big|\big|3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big|}$
$=\frac{6-12-30}{\sqrt{4+9+36}\sqrt{9+16+25}}$
$=\frac{-36}{\sqrt{49}\sqrt{50}}$
$=\frac{-36}{35\sqrt{2}}$
Rationalising the result, we get
$\cos\theta=-\frac{18\sqrt{2}}{35}$
$\therefore\theta=\cos^{-1}\Big(-\frac{18\sqrt{2}}{35}\Big)$
Thus, the angle between the given vectors measures $\cos^{-1}\Big(-\frac{18\sqrt{2}}{35}\Big)$.
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