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

the vector equations of two lines L_1 and L_2 is given by L_1 :r= i−j+3k + λ(i−j +k) L_2 : r= 2i+aj + 6k + μ(2i + j + 3k) where a,λ,μ are real constants. given that L_1 and L_2 intersect find a. the value of the constant a. b. the position vector of the point of intersection between L_1 and L_2 c. the cosine of the acute angle between L_1 and L_2 please help

$${the}\:{vector}\:{equations}\:{of}\:{two}\:{lines}\:{L}_{\mathrm{1}} \:{and}\:{L}_{\mathrm{2}} \:{is}\:{given}\:{by} \\ $$$$\:{L}_{\mathrm{1}} :{r}=\:\boldsymbol{{i}}−\boldsymbol{{j}}+\mathrm{3}\boldsymbol{{k}}\:+\:\lambda\left(\boldsymbol{{i}}−\boldsymbol{{j}}\:+\boldsymbol{{k}}\right) \\ $$$${L}_{\mathrm{2}} \::\:{r}=\:\mathrm{2}\boldsymbol{{i}}+{a}\boldsymbol{{j}}\:+\:\mathrm{6}\boldsymbol{{k}}\:+\:\mu\left(\mathrm{2}\boldsymbol{{i}}\:+\:\boldsymbol{{j}}\:+\:\mathrm{3}\boldsymbol{{k}}\right) \\ $$$${where}\:{a},\lambda,\mu\:{are}\:{real}\:{constants}. \\ $$$${given}\:{that}\:{L}_{\mathrm{1}} \:{and}\:{L}_{\mathrm{2}} \:{intersect}\:{find} \\ $$$${a}.\:\:{the}\:{value}\:{of}\:{the}\:{constant}\:{a}. \\ $$$${b}.\:\:{the}\:{position}\:{vector}\:{of}\:{the}\:{point}\:{of}\: \\ $$$${intersection}\:{between}\:{L}_{\mathrm{1}} \:{and}\:{L}_{\mathrm{2}} \\ $$$${c}.\:{the}\:{cosine}\:{of}\:{the}\:{acute}\:{angle}\:{between}\:{L}_{\mathrm{1}} \:{and}\:{L}_{\mathrm{2}} \\ $$$${please}\:{help} \\ $$$$ \\ $$

Commented by Kunal12588 last updated on 07/Dec/19

are these your hw question class 12

$${are}\:{these}\:{your}\:{hw}\:{question}\:{class}\:\mathrm{12} \\ $$

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

haha no sir,my revision question for my country′s A level examination,i wanna comfirm my answers

$${haha}\:{no}\:{sir},{my}\:{revision}\:{question}\:{for}\:{my} \\ $$$${country}'{s}\:{A}\:{level}\:{examination},{i}\:{wanna}\:{comfirm}\:{my} \\ $$$${answers} \\ $$

Answered by Kunal12588 last updated on 07/Dec/19

L_1 and L_2 intersect ∴ shortest dist b/w L_1 &L_2 = 0 ∴∣(((b_1 ^→ ×b_2 ^→ )∙(a_2 ^→ −a_1 ^→ ))/(∣b_1 ^→ ×b_2 ^→ ∣))∣=0 ⇒ determinant (((2−1),(a+1),(6−3)),(( 1),(−1),( 1)),(( 2),( 1),( 3)))=0 ⇒ determinant ((1,(a+1),3),(1,(−1),1),(2,( 1),3))=0 ⇒1(−3−1)−(a+1)(3−2)+3(1+2)=0 ⇒−4−a−1+9=0 ⇒a=4

$${L}_{\mathrm{1}} \:{and}\:{L}_{\mathrm{2}} \:{intersect}\: \\ $$$$\therefore\:{shortest}\:{dist}\:{b}/{w}\:\:{L}_{\mathrm{1}} \&{L}_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\therefore\mid\frac{\left(\overset{\rightarrow} {{b}}_{\mathrm{1}} ×\overset{\rightarrow} {{b}}_{\mathrm{2}} \right)\centerdot\left(\overset{\rightarrow} {{a}}_{\mathrm{2}} −\overset{\rightarrow} {{a}}_{\mathrm{1}} \right)}{\mid\overset{\rightarrow} {{b}}_{\mathrm{1}} ×\overset{\rightarrow} {{b}}_{\mathrm{2}} \mid}\mid=\mathrm{0} \\ $$$$\Rightarrow\begin{vmatrix}{\mathrm{2}−\mathrm{1}}&{{a}+\mathrm{1}}&{\mathrm{6}−\mathrm{3}}\\{\:\:\:\:\mathrm{1}}&{−\mathrm{1}}&{\:\:\:\mathrm{1}}\\{\:\:\:\:\mathrm{2}}&{\:\:\:\mathrm{1}}&{\:\:\:\mathrm{3}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow\begin{vmatrix}{\mathrm{1}}&{{a}+\mathrm{1}}&{\mathrm{3}}\\{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{1}}\\{\mathrm{2}}&{\:\:\:\mathrm{1}}&{\mathrm{3}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}\left(−\mathrm{3}−\mathrm{1}\right)−\left({a}+\mathrm{1}\right)\left(\mathrm{3}−\mathrm{2}\right)+\mathrm{3}\left(\mathrm{1}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow−\mathrm{4}−{a}−\mathrm{1}+\mathrm{9}=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{4} \\ $$

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

thank you

$${thank}\:{you} \\ $$

Answered by Kunal12588 last updated on 07/Dec/19

Another way r of L_1 =r of L_2 (1+λ)i+(−1−λ)j+(3+λ)k=(2+2μ)i+(a+μ)j+(6+3μ)k 1+λ=2+2μ −1−λ=a+μ 3+λ=6+3μ ⇒λ−2μ=1 ⇒λ+μ=−1−a ⇒λ−3μ=3 ⇒−2μ+3μ=1−3 ⇒μ=−2 ⇒λ+4=1⇒λ=−3 ⇒−2−3=−1−a ⇒a=4

$${Another}\:{way} \\ $$$${r}\:{of}\:{L}_{\mathrm{1}} ={r}\:{of}\:{L}_{\mathrm{2}} \\ $$$$\left(\mathrm{1}+\lambda\right){i}+\left(−\mathrm{1}−\lambda\right){j}+\left(\mathrm{3}+\lambda\right){k}=\left(\mathrm{2}+\mathrm{2}\mu\right){i}+\left({a}+\mu\right){j}+\left(\mathrm{6}+\mathrm{3}\mu\right){k} \\ $$$$\mathrm{1}+\lambda=\mathrm{2}+\mathrm{2}\mu\:\:\:\:\:\:−\mathrm{1}−\lambda={a}+\mu\:\:\:\:\:\:\:\:\:\:\mathrm{3}+\lambda=\mathrm{6}+\mathrm{3}\mu \\ $$$$\Rightarrow\lambda−\mathrm{2}\mu=\mathrm{1}\:\:\:\:\:\:\:\Rightarrow\lambda+\mu=−\mathrm{1}−{a}\:\:\:\:\Rightarrow\lambda−\mathrm{3}\mu=\mathrm{3} \\ $$$$\Rightarrow−\mathrm{2}\mu+\mathrm{3}\mu=\mathrm{1}−\mathrm{3} \\ $$$$\Rightarrow\mu=−\mathrm{2}\:\:\Rightarrow\lambda+\mathrm{4}=\mathrm{1}\Rightarrow\lambda=−\mathrm{3} \\ $$$$\Rightarrow−\mathrm{2}−\mathrm{3}=−\mathrm{1}−{a} \\ $$$$\Rightarrow{a}=\mathrm{4} \\ $$

Answered by Kunal12588 last updated on 07/Dec/19

r=i−j+3k+λ(i−j+k) a^→ =i−j+3k−3(i−j+k) {λ=−3 for intersection} =(1−3)i−(1−3)j+(3−3)k =−2i+2j

$${r}={i}−{j}+\mathrm{3}{k}+\lambda\left({i}−{j}+{k}\right) \\ $$$$\overset{\rightarrow} {{a}}={i}−{j}+\mathrm{3}{k}−\mathrm{3}\left({i}−{j}+{k}\right)\:\:\:\left\{\lambda=−\mathrm{3}\:{for}\:{intersection}\right\} \\ $$$$=\left(\mathrm{1}−\mathrm{3}\right){i}−\left(\mathrm{1}−\mathrm{3}\right){j}+\left(\mathrm{3}−\mathrm{3}\right){k} \\ $$$$=−\mathrm{2}{i}+\mathrm{2}{j} \\ $$

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

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by Kunal12588 last updated on 07/Dec/19

angle between L_1 &L_2 cos θ = ∣((b_1 ^→ ∙b_2 ^→ )/(∣b_1 ^→ ∣∣b_2 ^→ ∣))∣=∣(((i−j+k)∙(2i+j+3k))/((√(1+1+1))(√(4+1+9))))∣ ⇒cos θ =∣((2−1+3)/((√3)(√(14))))∣=(4/(√(42)))

$${angle}\:{between}\:{L}_{\mathrm{1}} \&{L}_{\mathrm{2}} \\ $$$${cos}\:\theta\:=\:\mid\frac{\overset{\rightarrow} {{b}}_{\mathrm{1}} \centerdot\overset{\rightarrow} {{b}}_{\mathrm{2}} }{\mid\overset{\rightarrow} {{b}}_{\mathrm{1}} \mid\mid\overset{\rightarrow} {{b}}_{\mathrm{2}} \mid}\mid=\mid\frac{\left({i}−{j}+{k}\right)\centerdot\left(\mathrm{2}{i}+{j}+\mathrm{3}{k}\right)}{\sqrt{\mathrm{1}+\mathrm{1}+\mathrm{1}}\sqrt{\mathrm{4}+\mathrm{1}+\mathrm{9}}}\mid \\ $$$$\Rightarrow{cos}\:\theta\:=\mid\frac{\mathrm{2}−\mathrm{1}+\mathrm{3}}{\sqrt{\mathrm{3}}\sqrt{\mathrm{14}}}\mid=\frac{\mathrm{4}}{\sqrt{\mathrm{42}}} \\ $$

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

thanks sir

$${thanks}\:{sir} \\ $$

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