Question
Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.
✓
Answer
The vectors, represented by these are
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
and $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
Let, $\theta$ be the angle between the lines,
then,
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\Big|\vec{\text{a}}\Big|\Big|\vec{\text{b}}\Big|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{(2)^2+(3)^2+(6)^2}\sqrt{(1)^2+(2)^2+(2)^2}}$
$=\frac{(2)(1)+(3)(2)+(6)(2)}{\sqrt{4+9+36}\sqrt{1+4+4}}$
$=\frac{2+6+12}{\sqrt{49}\sqrt{9}}$
$=\frac{20}{7\times3}$
$\cos\theta=\frac{20}{21}$
$\theta=\cos^{-1}\Big(\frac{20}{21}\Big)$
Angle between the lines $=\cos^{-1}\Big(\frac{20}{21}\Big)$.
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
and $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
Let, $\theta$ be the angle between the lines,
then,
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\Big|\vec{\text{a}}\Big|\Big|\vec{\text{b}}\Big|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{(2)^2+(3)^2+(6)^2}\sqrt{(1)^2+(2)^2+(2)^2}}$
$=\frac{(2)(1)+(3)(2)+(6)(2)}{\sqrt{4+9+36}\sqrt{1+4+4}}$
$=\frac{2+6+12}{\sqrt{49}\sqrt{9}}$
$=\frac{20}{7\times3}$
$\cos\theta=\frac{20}{21}$
$\theta=\cos^{-1}\Big(\frac{20}{21}\Big)$
Angle between the lines $=\cos^{-1}\Big(\frac{20}{21}\Big)$.
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