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NCERT Class XII Mathematics Chapter - - Solutions

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Question : 16 of 78
Marks: +1, -0
Find the angle between the pairs of lines:
r⃗=3i^+j^−2k^\vec{r}= \hat{3i}+ \hat{j} -\hat{2k} +λ(i^−j^−2k^)+\lambda( \hat{i} - \hat{j} -\hat{2k}) and r⃗=2i^−j^−56k^\vec{r}= \hat{2i}-\hat{j}-\hat{56k} +μ(3i^−5j^−4k^)+\mu( \hat{3i}-\hat{5j}-\hat{4k})
Solution:  
Here, b⃗1=i^−j^−2k^\vec{b}_1 = \hat{i} - \hat{j} -\hat{2k} and b⃗2=3i^−5j^−4k^\vec{b}_2 = \hat{3i} - \hat{5j} -\hat{4k}
Let θ be the angle between the given lines, then
cos⁡θ=∣b⃗1⋅b⃗2∣b⃗1∣∣b⃗2∣∣\cos \theta = \left| \frac{ \vec{b}_1 \cdot \vec{b}_2 }{ |\vec{b}_1| |\vec{b}_2| } \right|
=∣1×3+(−1)×(−5)+(−2)×(−4)∣(1)2+(−1)2+(−2)232+(−5)2+(−4)2= \frac{ |1 \times 3 + (-1) \times (-5) + (-2) \times (-4)| }{ \sqrt{ (1)^2+(-1)^2+(-2)^2 } \sqrt{ 3^2+(-5)^2+(-4)^2 } }
=16103=853= \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}
⇒θ=cos⁡−1(853)\Rightarrow \theta = \cos^{-1}\left( \frac{8}{5\sqrt{3}} \right)
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