Find the angle between the pair of lines r=2 i-5 j+k+ (3 i+2 j+6 k) and r=7 i-6 k+ (i+2 j+2 k)

We know that If is the acute angle between r = a_ 1 + b_ 1 and r = a_ 2 + b_ 2 , then = | c b_ 1 b_ 2 |b_ 1 ||b_ 2 | | .......(i) Given that, r =2 i -5 j + k + (3 i +2 j +6 k ) and r =7 i -6 k + ( i +2 j +2 k ) Here, b_ 1 =3 i +2 j +6 k and b_ 2 = i +2 j +2 k So, from (i), we have = | (3 i+2 j+6 k) (i+2 j+2 k) |3 i+2 j+6 k| |i+2 j+2 k| | ......(ii) | a i + b j +c k |= a^ 2 +b^ 2 +c^ 2 |3 i+2 j+6 k| = 3^ 2 +2^ 2 +6^ 2 =9+4+36=49=7 And |i+2 j+2 k|= 1^ 2 +2^ 2 +2^ 2 =1+4+4=9=3 Now, (a_ 1 i+b_ 1 j+c_ 1 k ) (a_ 2 i+b_ 2 j+c_ 2 k )=a_ 1 a_ 2 +b_ 1 b_ 2 +c_ 1 c_ 2 (3 i+2 j+6 k) (i+2 j+2 k) = 3 1+2 2+6 2=3+4+12=19 ⇒ By (ii), we have =19 7 3 =19 21 = ^ -1 (19 21 )

Find the angle between the pair of lines r=2 i-5 j+k+ (3 i+2 j+6 …

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Question

Find the angle between the pair of lines
$\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})$ and $\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$

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Answer

We know that
If $\theta$ is the acute angle between $\vec{\mathrm{r}}=\vec{\mathrm{a}_{1}}+\lambda \vec{\mathrm{b}_{1}}$ and $\vec{\mathrm{r}}=\vec{\mathrm{a}_{2}}+\mu \vec{\mathrm{b}_{2}}$ , then
$\cos \theta=\left|\begin{array}{c} \frac {{\vec{b}_{1} \cdot \vec{b}_{2}}} {{|\vec {b}_{1}||\vec {b}_{2}}|} \end{array}\right|$ .......(i)
Given that, $\vec{\mathrm{r}}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+\hat{\mathrm{k}}+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$ and $\vec{\mathrm{r}}=7 \hat{\mathrm{i}}-6 \hat{\mathrm{k}}+\mu(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$
Here, $\vec{b}_{1}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ and $\vec{b}_{2}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
So, from (i), we have
$\cos \theta=\left|\frac{(3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})}{|\hat{3} \hat{i}+2 \hat{j}+6 \hat{k}| \cdot|\hat{i}+2 \hat{j}+2 \hat{k}|}\right|$ ......(ii)
$\Rightarrow\left|{a}{\hat{i}}+{\mathrm{b} \hat{\mathrm{j}}}+\mathrm{c} \hat{\mathrm{k}}\right|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}$
$\Rightarrow |3 \hat{i}+2 \hat{j}+6 \hat{k}|$ = $\sqrt{3^{2}+2^{2}+6^{2}}=\sqrt{9+4+36}=\sqrt{49}=7$
And $|\hat{i}+2 \hat{j}+2 \hat{k}|=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{1+4+4}=\sqrt{9}=3$
Now, $\left(a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}\right) \cdot\left(a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}\right)=a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}$
$\Rightarrow (3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})$ = $3 \times 1+2 \times 2+6 \times 2=3+4+12=19$
⇒ By (ii), we have
$\Rightarrow \cos \theta=\frac{19}{7 \times 3}=\frac{19}{21}$
$\Rightarrow \theta=\cos ^{-1}\left(\frac{19}{21}\right)$

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