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Q.Find the shortest distance between the lines : r⃗=(i^+2j^+k^)+λ(i^−j^+k^)\vec r = (\hat i + 2\hat j + \hat k) + \lambda(\hat i - \hat j + \hat k) and r⃗=(2i^−j^−k^)+μ(2i^+j^+2k^)\vec r = (2\hat i - \hat j - \hat k) + \mu(2\hat i + \hat j + 2\hat k) OR Find the angle between the planes whose vector equations are : r⃗⋅(2i^+2j^−3k^)=5\vec r \cdot (2\hat i + 2\hat j - 3\hat k) = 5 and r⃗⋅(3i^−3j^+5k^)=3\vec r \cdot (3\hat i - 3\hat j + 5\hat k) = 3

Jammu Kashmir JkboseJKBOSE Class 12 Annual Regular Examination 2020Subjective· 6mImportance★★★★★
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Shortest distance between skew lines uses ∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}; angle between planes uses the normals' dot product.

Main part. a⃗1=i^+2j^+k^\vec a_1=\hat i+2\hat j+\hat k, b⃗1=i^−j^+k^\vec b_1=\hat i-\hat j+\hat k; a⃗2=2i^−j^−k^\vec a_2=2\hat i-\hat j-\hat k, b⃗2=2i^+j^+2k^\vec b_2=2\hat i+\hat j+2\hat k.

a⃗2−a⃗1=i^−3j^−2k^\vec a_2-\vec a_1 = \hat i-3\hat j-2\hat k.

b⃗1×b⃗2=∣i^j^k^1−11212∣=i^(−2−1)−j^(2−2)+k^(1+2)=−3i^+0j^+3k^\vec b_1\times\vec b_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix} = \hat i(-2-1) - \hat j(2-2) + \hat k(1+2) = -3\hat i+0\hat j+3\hat k

∣b⃗1×b⃗2∣=9+0+9=32|\vec b_1\times\vec b_2| = \sqrt{9+0+9}=3\sqrt2

(a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(1)(−3)+(−3)(0)+(−2)(3)=−9(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2) = (1)(-3)+(-3)(0)+(-2)(3) = -9

Shortest distance =∣−9∣32=932=32=322=\dfrac{|-9|}{3\sqrt2}=\dfrac{9}{3\sqrt2}=\dfrac{3}{\sqrt2}=\dfrac{3\sqrt2}{2} units.

OR (alternative part). Normals: n⃗1=2i^+2j^−3k^\vec n_1=2\hat i+2\hat j-3\hat k, n⃗2=3i^−3j^+5k^\vec n_2=3\hat i-3\hat j+5\hat k.

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