Find the angle between the lines = 3 - 2 + 6 + (2 + + 2 ) and = (2 - 5 ) + (6 + 3 + 2 ) As we know, = 1 2 | | | | 2 | where, is the angle between the lines 1 + 1 and 2 + 2 .

It is given that, = 3 - 2 + 6 + (2 + + 2 ) and = (2 - 5 ) + (6 + 3 + 2 ) where, 1 = 3 - 2 + 6 , 1 = 2 + + 2 and 2 = 2 - 5 , 2 = 6 + 3 + 2 If is angle between the lines, then = 1 2 | | 1 | | 2 | = + + 2 ) (6 + 3 + 2 )| |2 + + 2 || + 3 + 2 | = 9 = 21 = - 1 21

Three Dimensional Geometry — Class 12 Maths Solution

exemplar sa SA NCERT,Exemp.Q.4,Page.235

Question

Find the angle between the lines
$\overrightarrow {\rm{r}} = 3\widehat {\rm{i}} - 2\widehat {\rm{j}} + 6\widehat {\rm{k}} + \lambda (2\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}})$

and $\overrightarrow {\rm{r}} = (2\widehat {\rm{j}} - 5\widehat {\rm{k}}) + \mu (6\widehat {\rm{i}} + 3\widehat {\rm{j}} + 2\widehat {\rm{k}})$

As we know, $\cos \theta = \frac{{\left| {{{\overrightarrow {\rm{b}} }_1} \cdot {{\overrightarrow {\rm{b}} }_2}} \right|}}{{|\overrightarrow {\rm{b}} | \cdot \left| {{{\overrightarrow {\rm{b}} }_2}} \right|}}$

where, $\theta$ is the angle between the lines ${\overrightarrow {\rm{a}} _1} + \lambda {\overrightarrow {\rm{b}} _1}$ and $\overrightarrow {{{\rm{a}}_2}} + \mu \overrightarrow {{{\rm{b}}_2}}$.

Step-by-step Solution

It is given that, $\overrightarrow {\rm{r}} = 3\widehat {\rm{i}} - 2\widehat {\rm{j}} + 6\widehat {\rm{k}} + \lambda (2\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}})$

and $\overrightarrow {\rm{r}} = (2\widehat {\rm{j}} - 5\widehat {\rm{k}}) + \mu (6\widehat {\rm{i}} + 3\widehat {\rm{j}} + 2\widehat {\rm{k}})$

where, $\overrightarrow {{{\rm{a}}_1}} = 3\widehat {\rm{i}} - 2\widehat {\rm{j}} + 6\widehat {\rm{k}},$ $\overrightarrow {{{\rm{b}}_1}} = 2\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}}$

and $\overrightarrow {{{\rm{a}}_2}} = 2\widehat {\rm{j}} - 5\widehat {\rm{k}},$ $\overrightarrow {{{\rm{b}}_2}} = 6\widehat {\rm{i}} + 3\widehat {\rm{j}} + 2\widehat {\rm{k}}$

If $\theta$ is angle between the lines, then

$\cos \theta = \frac{{\left| {{{\overrightarrow {\rm{b}} }_1} \cdot {{\overrightarrow {\rm{b}} }_2}} \right|}}{{\left| {{{\overrightarrow {\rm{b}} }_1}} \right| \cdot \mid \left| {\overrightarrow {{{\rm{b}}_2}} } \right|}}$

$= \frac{{|(2\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}}) \cdot (6\widehat {\rm{i}} + 3\widehat {\rm{j}} + 2\widehat {\rm{k}})|}}{{|2\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}}||\widehat {\rm{i}} + 3\widehat {\rm{j}} + 2\widehat {\rm{k}}|}}$

$= \frac{{|12 + 3 + 4|}}{{\sqrt 9 \sqrt {49} }} = \frac{{19}}{{21}}$
$\therefore$ $\theta = {\cos ^{ - 1}}\frac{{19}}{{21}}$

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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Three Dimensional Geometry. Curated by Sachin Sharma. Free for all students.