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Q.Find the shortest distance between the lines r⃗=6i^+2j^+2k^+λ(i^−2j^+2k^)\vec r = 6\hat i + 2\hat j + 2\hat k + \lambda(\hat i - 2\hat j + 2\hat k) and r⃗=−4i^−k^+μ(3i^−2j^−2k^)\vec r = -4\hat i - \hat k + \mu(3\hat i - 2\hat j - 2\hat k).

(OR)
Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2i^−j^+4k^2\hat i - \hat j + 4\hat k and is in the direction i^+2j^−k^\hat i + 2\hat j - \hat k.
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2025Subjective· 5mImportance★★★★★
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Use the skew-line shortest-distance formula d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣d=\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}.

a⃗1=(6,2,2)\vec a_1=(6,2,2), b⃗1=(1,−2,2)\vec b_1=(1,-2,2); a⃗2=(−4,0,−1)\vec a_2=(-4,0,-1), b⃗2=(3,−2,−2)\vec b_2=(3,-2,-2).

a⃗2−a⃗1=(−10,−2,−3)\vec a_2-\vec a_1=(-10,-2,-3).

b⃗1×b⃗2=∣i^j^k^1−223−2−2∣=i^[(−2)(−2)−(2)(−2)]−j^[(1)(−2)−(2)(3)]+k^[(1)(−2)−(−2)(3)]\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-2&2\\3&-2&-2\end{vmatrix}=\hat i[(-2)(-2)-(2)(-2)]-\hat j[(1)(-2)-(2)(3)]+\hat k[(1)(-2)-(-2)(3)]

=i^(4+4)−j^(−2−6)+k^(−2+6)=8i^+8j^+4k^=\hat i(4+4)-\hat j(-2-6)+\hat k(-2+6)=8\hat i+8\hat j+4\hat k

∣b⃗1×b⃗2∣=64+64+16=144=12|\vec b_1\times\vec b_2|=\sqrt{64+64+16}=\sqrt{144}=12

(a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(−10)(8)+(−2)(8)+(−3)(4)=−80−16−12=−108(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=(-10)(8)+(-2)(8)+(-3)(4)=-80-16-12=-108

d=∣−108∣12=10812=9d=\dfrac{|-108|}{12}=\dfrac{108}{12}=9

Shortest distance =9=9 units.

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