the-vector-equations-of-two-lines-L-1-and-L-2-is-given-by-L-1-r-i-j-3k-i-j-k-L-2-r-2i-aj-6k-2i-j-3k-where-a-are-real-constants-given-that-L-1-and-L-2-intersect-find-a-

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Question Number 75101 by Rio Michael last updated on 07/Dec/19

$$thevectorequationsoftwolines{L}_{1}and{L}_{2}isgivenby$$$$L_1:r=\mathit{i}-\mathit{j}+3\mathit{k}+\lambda (\mathit{i}-\mathit{j}+\mathit{k})$$$$L_2:r=2\mathit{i}+a\mathit{j}+6\mathit{k}+\mu (2\mathit{i}+\mathit{j}+3\mathit{k})$$$$wherea,\lambda ,\mu arerealconstants.$$$$giventhat{L}_{1}and{L}_{2}intersectfind$$$$a.thevalueoftheconstanta.$$$$b.thepositionvectorofthepointof$$$$intersectionbetween{L}_{1}and{L}_2$$$$c.thecosineoftheacuteanglebetween{L}_{1}and{L}_2$$$$pleasehelp$$$$$$

Commented by Kunal12588 last updated on 07/Dec/19

$$aretheseyourhwquestionclass12$$

Commented by Rio Michael last updated on 07/Dec/19

$$hahanosir,myrevisionquestionformy$$$${country}^{\prime }sAlevelexamination,iwannacomfirmmy$$$$answers$$

Answered by Kunal12588 last updated on 07/Dec/19

$$L_{1}and{L}_{2}intersect$$$$\therefore shortestdistb/w{L}_1\mathrm{\&}{L}_2=0$$$$\therefore \mid \frac{\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right)\cdot ({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)}{\mid {\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\mid }\mid =0$$$$\Rightarrow \left|\begin{array}{ccc}2-1& a+1& 6-3\\ 1& -1& 1\\ 2& 1& 3\end{array}\right|=0$$$$\Rightarrow \left|\begin{array}{ccc}1& a+1& 3\\ 1& -1& 1\\ 2& 1& 3\end{array}\right|=0$$$$\Rightarrow 1(-3-1)-(a+1)(3-2)+3(1+2)=0$$$$\Rightarrow -4-a-1+9=0$$$$\Rightarrow a=4$$

Commented by peter frank last updated on 07/Dec/19

$$thankyou$$

Answered by Kunal12588 last updated on 07/Dec/19

$$Anotherway$$$$rof{L}_1=rof{L}_2$$$$(1+\lambda )i+(-1-\lambda )j+(3+\lambda )k=(2+2\mu )i+(a+\mu )j+(6+3\mu )k$$$$1+\lambda =2+2\mu -1-\lambda =a+\mu 3+\lambda =6+3\mu$$$$\Rightarrow \lambda -2\mu =1\Rightarrow \lambda +\mu =-1-a\Rightarrow \lambda -3\mu =3$$$$\Rightarrow -2\mu +3\mu =1-3$$$$\Rightarrow \mu =-2\Rightarrow \lambda +4=1\Rightarrow \lambda =-3$$$$\Rightarrow -2-3=-1-a$$$$\Rightarrow a=4$$

Answered by Kunal12588 last updated on 07/Dec/19

$$r=i-j+3k+\lambda (i-j+k)$$$$\overrightarrow{a}=i-j+3k-3(i-j+k)\{\lambda =-3forintersection\}$$$$=(1-3)i-(1-3)j+(3-3)k$$$$=-2i+2j$$

Commented by Rio Michael last updated on 07/Dec/19

$$thankyousir$$

Answered by Kunal12588 last updated on 07/Dec/19

$$anglebetween{L}_1\mathrm{\&}{L}_2$$$$cos\theta =\mid \frac{{\overrightarrow{b}}_1\cdot {\overrightarrow{b}}_2}{\mid {\overrightarrow{b}}_1\mid \mid {\overrightarrow{b}}_2\mid }\mid =\mid \frac{(i-j+k)\cdot (2i+j+3k)}{\sqrt{1+1+1}\sqrt{4+1+9}}\mid$$$$\Rightarrow cos\theta =\mid \frac{2-1+3}{\sqrt{3}\sqrt{14}}\mid =\frac{4}{\sqrt{42}}$$

Commented by Rio Michael last updated on 07/Dec/19

$$thankssir$$

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