# 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}).

Category: QuestionsFind 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}).
Editor">Editor Staff asked 11 months ago

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
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options
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Editor">Editor Staff answered 11 months ago

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}}