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Q.Find the shortest distance between the lines r⃗=(i^+j^)+λ(2i^−j^+k^)\vec{r} = (\hat{i}+\hat{j}) + \lambda(2\hat{i}-\hat{j}+\hat{k}) and r⃗=(2i^+j^−k^)+μ(3i^−5j^+2k^)\vec{r} = (2\hat{i}+\hat{j}-\hat{k}) + \mu(3\hat{i}-5\hat{j}+2\hat{k}).

(OR)
Find the coordinates of the point where the line through the points A(3,4,1)A(3,4,1) and B(5,1,6)B(5,1,6) crosses the XY-plane.
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2023Subjective· 4mImportance★★★★★
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Main part: apply the skew-lines shortest-distance formula ∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣\left|\dfrac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|. OR part: parametrise the line through A, B and set z=0z=0.

Main part. Line 1: a⃗1=i^+j^\vec a_1=\hat i+\hat j, b⃗1=2i^−j^+k^\vec b_1=2\hat i-\hat j+\hat k. Line 2: a⃗2=2i^+j^−k^\vec a_2=2\hat i+\hat j-\hat k, b⃗2=3i^−5j^+2k^\vec b_2=3\hat i-5\hat j+2\hat k.

a⃗2−a⃗1=i^+0j^−k^=(1,0,−1)\vec a_2-\vec a_1=\hat i+0\hat j-\hat k=(1,0,-1).

b⃗1×b⃗2=∣i^j^k^2−113−52∣=i^((−1)(2)−(1)(−5))−j^((2)(2)−(1)(3))+k^((2)(−5)−(−1)(3))\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&1\\3&-5&2\end{vmatrix}=\hat i((-1)(2)-(1)(-5))-\hat j((2)(2)-(1)(3))+\hat k((2)(-5)-(-1)(3))

=i^(−2+5)−j^(4−3)+k^(−10+3)=3i^−j^−7k^=\hat i(-2+5)-\hat j(4-3)+\hat k(-10+3)=3\hat i-\hat j-7\hat k.

∣b⃗1×b⃗2∣=9+1+49=59|\vec b_1\times\vec b_2|=\sqrt{9+1+49}=\sqrt{59}.

(a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(1)(3)+(0)(−1)+(−1)(−7)=3+0+7=10(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=(1)(3)+(0)(-1)+(-1)(-7)=3+0+7=10.

Shortest distance =∣10∣59=1059=\dfrac{|10|}{\sqrt{59}}=\dfrac{10}{\sqrt{59}}.

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