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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 vector equation of the line passing through the point (1,2,−4)(1, 2, -4) and perpendicular to the two lines x−83=y+19−16=z−107\dfrac{x-8}{3} = \dfrac{y+19}{-16} = \dfrac{z-10}{7} and x−153=y−298=z−5−5\dfrac{x-15}{3} = \dfrac{y-29}{8} = \dfrac{z-5}{-5}.

Haryana BsehBSEH Intermediate Board 2026Subjective· 5mImportance★★★★★
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Use the skew-line shortest-distance formula d=∣(a2⃗−a1⃗)⋅(b1⃗×b2⃗)∣b1⃗×b2⃗∣∣d=\left|\dfrac{(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})}{|\vec{b_1}\times\vec{b_2}|}\right|.

Main question: Lines: r⃗=(6,2,2)+λ(1,−2,2)\vec{r}=(6,2,2)+\lambda(1,-2,2) and r⃗=(−4,0,−1)+μ(3,−2,−2)\vec{r}=(-4,0,-1)+\mu(3,-2,-2).

a2⃗−a1⃗=(−10,−2,−3)\vec{a_2}-\vec{a_1}=(-10,-2,-3)

b1⃗×b2⃗=∣i^j^k^1−223−2−2∣=i^(4+4)−j^(−2−6)+k^(−2+6)=(8,8,4)\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}(4+4)-\hat{j}(-2-6)+\hat{k}(-2+6)=(8,8,4)

∣b1⃗×b2⃗∣=64+64+16=144=12|\vec{b_1}\times\vec{b_2}|=\sqrt{64+64+16}=\sqrt{144}=12

(a2⃗−a1⃗)⋅(b1⃗×b2⃗)=−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=9d=\dfrac{|-108|}{12}=9

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