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Q.Find the shortest distance between the lines r⃗=(3i^+3j^−5k^)+μ(2i^+3j^+6k^)\vec{r}=(3\hat{i}+3\hat{j}-5\hat{k})+\mu(2\hat{i}+3\hat{j}+6\hat{k}) and r⃗=(i^+2j^−4k^)+λ(2i^−3j^+6k^)\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+\lambda(2\hat{i}-3\hat{j}+6\hat{k}). OR

(i) Solve: dydx+ytan⁡x=y2sec⁡x\dfrac{dy}{dx}+y\tan x=y^2\sec x.
(ii) Show that for any two vectors a⃗\vec{a} and b⃗\vec{b} it is always true that ∣a⃗+b⃗∣≤∣a⃗∣+∣b⃗∣|\vec{a}+\vec{b}|\le|\vec{a}|+|\vec{b}|.
Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2020Subjective· 8mImportance★★★★★
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Use 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|}; the cross product is (36,0,−12)(36,0,-12), giving d=841210=710d=\frac{84}{12\sqrt{10}}=\frac{7}{\sqrt{10}}.

Concept. For skew lines r⃗=a⃗1+μb⃗1\vec r=\vec a_1+\mu\vec b_1 and r⃗=a⃗2+λb⃗2\vec r=\vec a_2+\lambda\vec b_2, the shortest distance is

d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣.d=\frac{\big|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)\big|}{|\vec b_1\times\vec b_2|}.

Here a⃗1=(3,3,−5), b⃗1=(2,3,6), a⃗2=(1,2,−4), b⃗2=(2,−3,6)\vec a_1=(3,3,-5),\ \vec b_1=(2,3,6),\ \vec a_2=(1,2,-4),\ \vec b_2=(2,-3,6).

Step 1 — cross product.

b⃗1×b⃗2=∣i^j^k^2362−36∣=i^(18+18)−j^(12−12)+k^(−6−6)=(36, 0, −12).\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&6\\2&-3&6\end{vmatrix}=\hat i(18+18)-\hat j(12-12)+\hat k(-6-6)=(36,\,0,\,-12).

∣b⃗1×b⃗2∣=362+02+122=1440=1210.|\vec b_1\times\vec b_2|=\sqrt{36^2+0^2+12^2}=\sqrt{1440}=12\sqrt{10}.

Step 2 — connecting vector and scalar triple product.

a⃗2−a⃗1=(−2,−1,1),(a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(−2)(36)+(−1)(0)+(1)(−12)=−84.\vec a_2-\vec a_1=(-2,-1,1),\qquad (\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=(-2)(36)+(-1)(0)+(1)(-12)=-84.

Step 3 — distance.

d=∣−84∣1210=841210=710=71010 units.d=\frac{|-84|}{12\sqrt{10}}=\frac{84}{12\sqrt{10}}=\frac{7}{\sqrt{10}}=\frac{7\sqrt{10}}{10}\ \text{units}.

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