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Q.Find the shortest distance between the lines: x+17=y+1−6=z+11\dfrac{x+1}{7} = \dfrac{y+1}{-6} = \dfrac{z+1}{1} and x−31=y−5−2=z−71\dfrac{x-3}{1} = \dfrac{y-5}{-2} = \dfrac{z-7}{1}. OR Find the distance of a point (2, 5, -3) from the plane: r⃗⋅(6i^−3j^+2k^)=4\vec{r}\cdot(6\hat{i} - 3\hat{j} + 2\hat{k}) = 4.

Jammu Kashmir JkboseJKBOSE Class 12 Annual Regular Examination 2018Subjective· 6mImportance★★★★★
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The shortest distance between two skew lines uses ∣(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|; the OR part uses the point-to-plane distance formula.

Part 1 — Shortest distance between x+17=y+1−6=z+11\dfrac{x+1}{7}=\dfrac{y+1}{-6}=\dfrac{z+1}{1} and x−31=y−5−2=z−71\dfrac{x-3}{1}=\dfrac{y-5}{-2}=\dfrac{z-7}{1}

Line 1 passes through a⃗1=(−1,−1,−1)\vec a_1=(-1,-1,-1) with direction b⃗1=(7,−6,1)\vec b_1=(7,-6,1).

Line 2 passes through a⃗2=(3,5,7)\vec a_2=(3,5,7) with direction b⃗2=(1,−2,1)\vec b_2=(1,-2,1).

a⃗2−a⃗1=(4,6,8)\vec a_2-\vec a_1=(4,6,8).

b⃗1×b⃗2=∣i^j^k^7−611−21∣=i^(−6−(−2))−j^(7−1)+k^(−14−(−6))=(−4,−6,−8)\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\7&-6&1\\1&-2&1\end{vmatrix}=\hat i(-6-(-2))-\hat j(7-1)+\hat k(-14-(-6))=(-4,-6,-8)

∣b⃗1×b⃗2∣=16+36+64=116=229|\vec b_1\times\vec b_2|=\sqrt{16+36+64}=\sqrt{116}=2\sqrt{29}

(a⃗2−a⃗1)⋅(b⃗1×b⃗2)=4(−4)+6(−6)+8(−8)=−16−36−64=−116(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=4(-4)+6(-6)+8(-8)=-16-36-64=-116

Shortest distance =∣−116∣229=116229=5829=229=\dfrac{|-116|}{2\sqrt{29}}=\dfrac{116}{2\sqrt{29}}=\dfrac{58}{\sqrt{29}}=2\sqrt{29} units. …

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