1.

Find the angle between the lines \(\vec{r}=2\hat{i}+6\hat{j}-\hat{k}+λ(\hat{i}-2\hat{j}+3\hat{k})\) and \(\vec{r}=4\hat{i}-7\hat{j}+3\hat{k}+μ(5\hat{i}-3\hat{j}+3\hat{k})\).(a) θ=\(cos^{-1}\frac{⁡20}{\sqrt{602}}\)(b) θ=\(cos^{-1}\frac{⁡20}{\sqrt{682}}\)(c) θ=\(cos^{-1}\frac{⁡8}{\sqrt{602}}\)(d) θ=\(cos^{-1}⁡\frac{14}{\sqrt{598}}\)I got this question in semester exam.I want to ask this question from Three Dimensional Geometry in portion Three Dimensional Geometry of Mathematics – Class 12

Answer»

Right choice is (a) θ=\(cos^{-1}\FRAC{⁡20}{\SQRT{602}}\)

The explanation is: If two lines have the equations \(\vec{r}=\vec{a_1}+λ\vec{b_1} \,and \,\vec{r}=\vec{a_2}+μ\vec{b_2}\)

Then, the angle between the two lines will be given by

cos⁡θ=\(\LEFT |\frac{\vec{b_1}.\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}\right |\)

=\(\left |\frac{(\hat{i}-2\hat{j}+3\hat{K}).(5\hat{i}-3\hat{j}+3\hat{k})}{\sqrt{1^2+(-2)^2+(3)^2).√(5^2+(-3)^2+3^2}}\right |\)

=\(\frac{5+6+9}{√14.√43}=\frac{20}{√602}\)

θ=\(cos^{-1}⁡\frac{20}{\sqrt{602}}\)


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