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Question: The angle between the straight line \(r = \left( {\hat i + 2\hat j + \hat k} \right) + \left( {\hat ...

The angle between the straight line r=(i^+2j^+k^)+(i^−j^+k^)r = \left( {\hat i + 2\hat j + \hat k} \right) + \left( {\hat i - \hat j + \hat k} \right) and the plane r⋅(2i^−j^+k^)=4r \cdot \left( {2\hat i - \hat j + \hat k} \right) = 4 is

A. sin⁡−1(223) B. sin⁡−1(26) C. sin⁡−1(23) D. sin⁡−1(23) E. sin⁡−1(23)  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) \\\
Explanation

Solution

Hint- In order to solve the problem use the relation giving the angle between the given line and a plane. If θ\theta is the angle between line r⃗=a⃗+λ(b⃗)\vec r = \vec a + \lambda \left( {\vec b} \right) and a plane r⃗⋅n⃗=d\vec r \cdot \vec n = d , in this condition sin⁡θ\sin \theta will be given as sin⁡θ=∣b⃗⋅n⃗∣b⃗∣∣n⃗∣∣\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=(i^+2j^+k^)+(i^−j^+k^)r = \left( {\hat i + 2\hat j + \hat k} \right) + \left( {\hat i - \hat j + \hat k} \right)
And the plane vector r⋅(2i^−j^+k^)=4r \cdot \left( {2\hat i - \hat j + \hat k} \right) = 4
As we know that If θ\theta is the angle between line r⃗=a⃗+λ(b⃗)\vec r = \vec a + \lambda \left( {\vec b} \right) and a plane r⃗⋅n⃗=d\vec r \cdot \vec n = d , in this condition sin⁡θ\sin \theta will be given as sin⁡θ=∣b⃗⋅n⃗∣b⃗∣∣n⃗∣∣ or θ=sin⁡−1∣b⃗⋅n⃗∣b⃗∣∣n⃗∣∣\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 θ\theta we will separately evaluate b⃗⋅n⃗ and ∣b⃗∣∣n⃗∣\vec b \cdot \vec n{\text{ and }}\left| {\vec b} \right|\left| {\vec n} \right|
Line vector is r=(i^+2j^+k^)+(i^−j^+k^)r = \left( {\hat i + 2\hat j + \hat k} \right) + \left( {\hat i - \hat j + \hat k} \right)
Comparing it with general line vector r⃗=a⃗+λ(b⃗)\vec r = \vec a + \lambda \left( {\vec b} \right)
a⃗=(i^+2j^+k^) and b⃗=(i^−j^+k^)\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⋅(2i^−j^+k^)=4r \cdot \left( {2\hat i - \hat j + \hat k} \right) = 4
Comparing it with general plane equation r⃗⋅n⃗=d\vec r \cdot \vec n = d
n⃗=(2i^−j^+k^) and d=4\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
(ai^+bj^+ck^)⋅(pi^+qj^+rk^)=ap+bq+cr\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 b⃗⋅n⃗\vec b \cdot \vec n

⇒b⃗⋅n⃗=(i^−j^+k^)⋅(2i^−j^+k^) ⇒b⃗⋅n⃗=1×2+(−1)×(−1)+1×1 ⇒b⃗⋅n⃗=2+1+1 ⇒b⃗⋅n⃗=4  \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
if a⃗=(xi^+yj^+zk^) then ∣a⃗∣=x2+y2+z2{\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 ∣b⃗∣∣n⃗∣\left| {\vec b} \right|\left| {\vec n} \right| , we proceed in the similar way
⇒∣b⃗∣∣n⃗∣=∣b⃗∣×∣n⃗∣ ⇒∣b⃗∣∣n⃗∣=∣(i^−j^+k^)∣×∣(2i^−j^+k^)∣ ⇒∣b⃗∣∣n⃗∣=(1)2+(−1)2+(1)2×(2)2+(−1)2+(1)2 ⇒∣b⃗∣∣n⃗∣=1+1+1×4+1+1 ⇒∣b⃗∣∣n⃗∣=3×6=18  \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
θ=sin⁡−1∣b⃗⋅n⃗∣b⃗∣∣n⃗∣∣\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
θ=sin⁡−1∣b⃗⋅n⃗∣b⃗∣∣n⃗∣∣ ⇒θ=sin⁡−1∣418∣ ⇒θ=sin⁡−1∣432∣=sin⁡−1∣223∣  \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(223){\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.