Examples On Vector Equations Of Planes Set-1

Go back to  'Vectors and 3-D Geometry'

Example – 51

Find the length of the perpendicular dropped from the point \(A(\vec a)\) onto the plane \(\vec r \cdot \hat n = d.\)

Solution:

Let O be the origin. The position vector of M can be written as

\[\begin{align}& {{\vec r}_M} = \overrightarrow {OA}  + \overrightarrow {AM}  \hfill \\\\&\quad\;   = \vec a - l\,\hat n\;\;\;\qquad where\;\;l = AM \hfill \\ \end{align} \]

Since M lies on the plane, we must have

\[\begin{align}&\qquad\;\;  {{\vec r}_M} \cdot \hat n = d \hfill \\\\&\Rightarrow \quad  (\vec a - l\,\hat n) \cdot \hat n = d \hfill \\\\&\Rightarrow \quad  \vec a \cdot \hat n - l = d \hfill \\\\& \Rightarrow  \quad l = \vec a \cdot \hat n - d \hfill \\ \end{align} \]

Thus, the length AM is  \(\left| {\vec a \cdot \hat n - d} \right|.\)

We could have evaluated this result straightaway by inspection using a slightly more elaborate diagram.

In particular, let us include the origin  \(O(\vec 0)\) in our diagram:

OP, the perpendicular from O onto the plane, is of length d since the equation of the plane is \(\vec r \cdot \hat n = d.\)

The component of OA along the extended line AM, say \(O{A_{AM}},\) will be \(\vec a \cdot \hat n.\) Thus,

\[\begin{align}&AM = \left| {O{A_{AM}} - OP} \right| \hfill \\\\&\qquad= \left| {\vec a \cdot \hat n - d} \right| \hfill \\ \end{align} \]

Example – 52

(a) Find the angle between the line \(\vec r = \vec a + \lambda \vec b\) and the plane \(\vec r \cdot \hat n = d.\)

(b) Find the angle between the planes \(\vec r \cdot {\hat n_1} = {d_1}\) and \(\vec r \cdot {\hat n_2} = {d_2}.\)

Solution:(a)

  \({{\theta '}}\)  is simply the angle between the vectors  \(\hat n\) and \(\vec b.\)

Thus, 

\[\cos {{\theta '}} = \frac{{\vec b \cdot \hat n}}{{\left| {\vec b} \right|\,\,\left| {\hat n} \right|}} = \hat b \cdot \hat n\]

The angle between the line and the plane is therefore given by

\[\begin{align}&\quad\quad\sin {{\theta }} = \cos {{\theta '}} = \hat b \cdot \hat n \hfill \\\\&\;\;\; \Rightarrow \quad {{\theta }} = {\sin ^{ - 1}}\left( {\hat b \cdot \hat n} \right) \hfill \\ \end{align} \]

(b) The angle \({\theta }\) between the two planes will simply be the angle between their normals. Thus,

\[\begin{align}&\qquad\quad\cos {{\theta }} = \frac{{{{\hat n}_1} \cdot {{\hat n}_2}}}{{\left| {{{\hat n}_1}} \right|\left| {{{\hat n}_2}} \right|}} = {{\hat n}_1}.{{\hat n}_2} \hfill \\& \Rightarrow \quad  {{\theta }} = {\cos ^{ - 1}}\left( {{{\hat n}_1} \cdot {{\hat n}_2}} \right) \hfill \\ \end{align} \]

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