Question

The angle between the straight line $r = \left( {\hat i + 2\hat j + \hat k} \right) + \left( {\hat i - \hat j + \hat k} \right)$ and the plane $r \cdot \left( {2\hat i - \hat j + \hat k} \right) = 4$ is
\[
  A.{\text{ }}{\sin ^{ - 1}}\left( {\dfrac{{2\sqrt 2 }}{3}} \right) \\
  B.{\text{ }}{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{6}} \right) \\
  C.{\text{ }}{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{3}} \right) \\
  D.{\text{ }}{\sin ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right) \\
  E.{\text{ }}{\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right) \\
 \]

Answer 1

Hint- In order to solve the problem use the relation giving the angle between the given line and a plane. If $\theta $ is the angle between line $\vec r = \vec a + \lambda \left( {\vec b} \right)$ and a plane \[\vec r \cdot \vec n = d\] , in this condition $\sin \theta $ will be given as $\sin \theta = \left| {\dfrac{{\vec b \cdot \vec n}}{{\left| {\vec b} \right|\left| {\vec n} \right|}}} \right|$

Complete step-by-step answer:

Given line vector $r = \left( {\hat i + 2\hat j + \hat k} \right) + \left( {\hat i - \hat j + \hat k} \right)$
And the plane vector $r \cdot \left( {2\hat i - \hat j + \hat k} \right) = 4$
As we know that If $\theta $ is the angle between line $\vec r = \vec a + \lambda \left( {\vec b} \right)$ and a plane \[\vec r \cdot \vec n = d\] , in this condition $\sin \theta $ will be given as $\sin \theta = \left| {\dfrac{{\vec b \cdot \vec n}}{{\left| {\vec b} \right|\left| {\vec n} \right|}}} \right|{\text{ or }}\theta = {\sin ^{ - 1}}\left| {\dfrac{{\vec b \cdot \vec n}}{{\left| {\vec b} \right|\left| {\vec n} \right|}}} \right|$
In order to use above property to find $\theta $ we will separately evaluate $\vec b \cdot \vec n{\text{ and }}\left| {\vec b} \right|\left| {\vec n} \right|$
Line vector is $r = \left( {\hat i + 2\hat j + \hat k} \right) + \left( {\hat i - \hat j + \hat k} \right)$
Comparing it with general line vector $\vec r = \vec a + \lambda \left( {\vec b} \right)$
$\vec a = \left( {\hat i + 2\hat j + \hat k} \right){\text{ and }}\vec b = \left( {\hat i - \hat j + \hat k} \right)$
Next plane vector is $r \cdot \left( {2\hat i - \hat j + \hat k} \right) = 4$
Comparing it with general plane equation \[\vec r \cdot \vec n = d\]
\[\vec n = \left( {2\hat i - \hat j + \hat k} \right){\text{ and }}d = 4\]
As we know that general dot product for two general vectors is carried out in following way
$\left( {a\hat i + b\hat j + c\hat k} \right) \cdot \left( {p\hat i + q\hat j + r\hat k} \right) = ap + bq + cr$
Similarly finding $\vec b \cdot \vec n$
\[
   \Rightarrow \vec b \cdot \vec n = \left( {\hat i - \hat j + \hat k} \right) \cdot \left( {2\hat i - \hat j + \hat k} \right) \\
   \Rightarrow \vec b \cdot \vec n = 1 \times 2 + \left( { - 1} \right) \times \left( { - 1} \right) + 1 \times 1 \\
   \Rightarrow \vec b \cdot \vec n = 2 + 1 + 1 \\
   \Rightarrow \vec b \cdot \vec n = 4 \\
 \]
Similarly we know mod of general vector is done in the following ways
${\text{if }}\vec a = \left( {x\hat i + y\hat j + z\hat k} \right){\text{ then }}\left| {\vec a} \right| = \sqrt {{x^2} + {y^2} + {z^2}} $
So for $\left| {\vec b} \right|\left| {\vec n} \right|$ , we proceed in the similar way
$
   \Rightarrow \left| {\vec b} \right|\left| {\vec n} \right| = \left| {\vec b} \right| \times \left| {\vec n} \right| \\
   \Rightarrow \left| {\vec b} \right|\left| {\vec n} \right| = \left| {\left( {\hat i - \hat j + \hat k} \right)} \right| \times \left| {\left( {2\hat i - \hat j + \hat k} \right)} \right| \\
   \Rightarrow \left| {\vec b} \right|\left| {\vec n} \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2}} \times \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2}} \\
   \Rightarrow \left| {\vec b} \right|\left| {\vec n} \right| = \sqrt {1 + 1 + 1} \times \sqrt {4 + 1 + 1} \\
   \Rightarrow \left| {\vec b} \right|\left| {\vec n} \right| = \sqrt 3 \times \sqrt 6 = \sqrt {18} \\
 $
As we know the formula for angle between them the line vector and the plane vector is
$\theta = {\sin ^{ - 1}}\left| {\dfrac{{\vec b \cdot \vec n}}{{\left| {\vec b} \right|\left| {\vec n} \right|}}} \right|$
So, now let us substitute the values in to the formula
$
  \theta = {\sin ^{ - 1}}\left| {\dfrac{{\vec b \cdot \vec n}}{{\left| {\vec b} \right|\left| {\vec n} \right|}}} \right| \\
   \Rightarrow \theta = {\sin ^{ - 1}}\left| {\dfrac{4}{{\sqrt {18} }}} \right| \\
   \Rightarrow \theta = {\sin ^{ - 1}}\left| {\dfrac{4}{{3\sqrt 2 }}} \right| = {\sin ^{ - 1}}\left| {\dfrac{{2\sqrt 2 }}{3}} \right| \\
 $
Hence, the angle between the line vector and the plane is ${\sin ^{ - 1}}\left( {\dfrac{{2\sqrt 2 }}{3}} \right)$
So, option A is the correct option.

Note- In order to solve such problems related to finding the angle between two figures whose coordinates are given in vector form, remembering the formula for the angle is very important. The problem can also be solved by basic steps of vectors and without the formula but that will be very lengthy. Students must remember the formula for dot product of vectors.