Examples On Vector Equations Of Planes Set-2

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Example – 53

It is physically obvious that two planes will intersect in a straight line.

Prove that the equation of the plane passing through the line of intersection of the planes

\begin{align}&{P_1} \equiv {{\vec r}_1} \cdot {{\vec n}_1} - {d_1} = 0 \hfill \\\\& {P_2} \equiv\; \vec r \cdot {{\vec n}_2} - {d_2}\; = 0 \hfill \\ \end{align}

can be written as $${P_1} + \lambda {P_2} = 0.$$

Solution: First of all, let us verify whether $${P_1} + \lambda {P_2} = 0$$ is indeed a plane:

\begin{align}&{P_1} + \lambda {P_2} \equiv (\vec r \cdot {{\vec n}_1} - {d_1}) + \lambda (\vec r \cdot {{\vec n}_2} - {d_2}) = 0 \hfill \\\\& \Rightarrow \quad \vec r \cdot \left( {{{\vec n}_1} + \lambda {{\vec n}_2}} \right) - \left( {{d_1} + \lambda {d_2}} \right) = 0\qquad\qquad\qquad...\left( 1 \right) \hfill \\ \end{align}

This of the form

$\vec r \cdot \vec n - d = 0$

and hence represents a plane.

For this plane to pass through the intersection line of P1 and P2, we must prove that every point lying on the intersection line of P1 and P2 must satisfy the equation $${P_1} + \lambda {P_2} = 0.$$ This is very straight forward. Assume $${\vec r_0}$$ to be any point lying on the intersection line P1 and P2 :

\begin{align}&\Rightarrow \quad {{\vec r}_0} \cdot {{\vec n}_1} - {d_1} = 0 \qquad , \qquad {{\vec r}_0} \cdot {{\vec n}_2} - {d_2} = 0 \hfill \\\\& \Rightarrow \quad {{\vec r}_0} \cdot {{\vec n}_1} - {d_1} + \lambda ({{\vec r}_0} \cdot {{\vec n}_2} - {d_2}) = 0 \hfill \\\\& \Rightarrow \quad {{\vec r}_0} \cdot \left( {{{\vec n}_1} + \lambda {{\vec n}_2}} \right) - \left( {{d_1} + \lambda {d_2}} \right) = 0 \hfill \\ \end{align}

This proves that $${\vec r_0}$$ satisfies (1). Thus, the stated assertion is correct.

Example – 54

Find the perpendicular distance of any corner of a cube of side a from a diagonal not passing through it.

Solution: Let us take the cube in the following configuration: Let us find the perpendicular distance d of O from AB. Using the result of Example 24, this equals

\begin{align}&d = \frac{{\left| {\overrightarrow {OA} \times \overrightarrow {OB} } \right|}}{{\left| {\overrightarrow {OA} - \overrightarrow {OB} } \right|}} \hfill \\\\&\;= \frac{{\left| {\hat i \times \left( {\hat j + \hat k} \right)} \right|}}{{\left| {\hat i - \hat j - \hat k} \right|}} \hfill \\\\&\; = \frac{{\left| {\hat k - \hat j} \right|}}{{\left| {\hat i - \hat j - \hat k} \right|}} \hfill \\&\; = \sqrt {\frac{2}{3}} \hfill \\ \end{align}

Example – 55

Find the position vector of the point of intersection of the three planes

$\vec r \cdot {\vec n_1} = {d_1},\,\,\,\vec r \cdot {\vec n_2} = {d_2},\,\,\,\vec r \cdot {\vec n_3} = {d_3}$

where $${\vec n_1},\,{\vec n_2}$$ and $${\vec n_3}$$ are non-coplanar vectors.

Solution: The condition that $${\vec n_1},\,{\vec n_2},{\vec n_3}$$ are non-coplanar will ensure that the three planes are guaranteed to intersect in in a point (think about this carefully). Let $${\vec r_0}$$ be the point of intersection of the three planes. We have,

${\vec r_0} \cdot {\vec n_1} = {d_1}\,\,\,,\,\,\,{\vec r_0} \cdot {\vec n_2} = {d_2},\,\,\,\,{\vec r_0} \cdot {\vec n_3} = {d_3}$

We must find a way to express $${\vec r_0}$$ in terms of the known vectors/ quantities. For this purpose, let us consider as the basis of our three dimensional space the vectors $$\left\{ {\left( {{{\vec n}_1} \times {{\vec n}_2}} \right)\,\,\,\,\left( {{{\vec n}_2} \times {{\vec n}_3}} \right)\,\,\,\,\,\left( {{{\vec n}_3} \times {{\vec n}_1}} \right)} \right\}.$$ This can be done since if $${\vec n_1},{\vec n_2},{\vec n_3}$$ are non-coplanar so are $${\vec n_1} \times {\vec n_2},\,\,{\vec n_2} \times {\vec n_3}$$ and $${\vec n_3} \times {\vec n_1}.$$ It will soon become clear why this should be done.

We can now write $${\vec r_0}$$ as

\begin{align}&\qquad\;\; {{\vec r}_0} = {\lambda _1}\left( {{{\vec n}_1} \times {{\vec n}_2}} \right) + {\lambda _2}\left( {{{\vec n}_2} \times {{\vec n}_3}} \right) + {\lambda _3}\left( {{{\vec n}_3} \times {{\vec n}_1}} \right)\,\,where\,\,{\lambda _1},\,{\lambda _2},\,{\lambda _3} \in \mathbb{R} \hfill \\\\& \Rightarrow \quad {{\vec r}_0} \cdot {{\vec n}_3} = {\lambda _1}\left[ {{{\vec n}_1}\,\,{{\vec n}_2}\,\,{{\vec n}_3}} \right] \quad \Rightarrow \quad {d_3} = {\lambda _1}\left[ {{{\vec n}_1}\,\,{{\vec n}_2}\,\,{{\vec n}_3}} \right] \hfill \\\\&\qquad\qquad\qquad\qquad\qquad\qquad\quad\Rightarrow \quad {\lambda _1} = \frac{{{d_3}}}{{\left[ {{{\vec n}_1}\,\,{{\vec n}_2}\,\,{{\vec n}_3}} \right]}} \hfill \\ \end{align}

$Similarly,{\lambda _2} = \frac{{{d_1}}}{{\left[ {{{\vec n}_1}\,\,{{\vec n}_2}\,\,{{\vec n}_3}} \right]}}\,\,{\text{ }}and\,\,{\lambda _3} = \frac{{{d_2}}}{{\left[ {{{\vec n}_1}\,\,{{\vec n}_2}\,\,{{\vec n}_3}} \right]}}$

Thus,

${\vec r_0} = \frac{1}{{\left[ {{{\vec n}_1}\,\,{{\vec n}_2}\,\,{{\vec n}_3}} \right]}}\,\,\left\{ {{d_3}\left( {{{\vec n}_1} \times {{\vec n}_2}} \right) + {d_1}\left( {{{\vec n}_2} \times {{\vec n}_3}} \right) + {d_2}\left( {{{\vec n}_3} \times {{\vec n}_1}} \right)} \right\}$

TRY YOURSELF - VI

Q. 1 Show that the external bisector of the angle A of a triangle ABC divides BC externally in the ratio AB : AC.

Q. 2 Find the equation of the plane containing the two parallel lines $$\vec r = \vec a + \lambda \vec b\,\,and\,\,\vec r = \vec \alpha + \mu \vec b$$ in non-parametric form.

Q. 3 Find the equation of the plane passing through the points $$A(2\hat i + \hat j + 3\hat k),\,\,B( - \hat i + 2\hat j + 4\hat k)\,\,and\,\,C(2\hat j + \hat k).$$ Find the position vector of the point D at which the line

$\vec r = (\hat i - \hat j + \hat k) + \lambda (2\hat i + \hat k)$

intersects the plane of triangle ABC.

Q. 4 In a  $$\Delta OAB,\,\,E$$ is the mid-point of OB and D is a point on AB such that AD : DB = 2 : 1. If OD and  AE intersect in P, prove that OP : OD = 3 : 5.

Q. 5 Points D, E, F divide the sides BC, CA and AB of respectively in the ratios 2 : 3, 1 : 2 and 3 : 1. Prove using vector methods that AD, BE and CF are concurrent.

Q. 6 Prove from first principles (that is, without using any other results) that the angle bisectors in any triangle are concurrent.

Q. 7 If one diagonal of a quadrilateral bisects the other, prove that it divides the quadrilateral into two triangles of equal areas.

Q. 8 In a triangle ABC, E and F are the mid -points of AC and AB respectively. CP is drawn parallel to AB to meet BE produced in P. Show that

${\text{area }}(\Delta FEP) = {\text{area }}(\Delta FCE) = \frac{1}{4}{\text{area }}(\Delta ABC)$

Q. 9 In a tetrahedron, the four lines joining the vertices of a tetrahedron to the centroids of the opposite faces are concurrent. (We’ve already proved this earlier).

Prove that the point of concurrency divides each of the four lines in the ratio 3 : 1.

Q. 10 Lines L1and L2 are drawn through  $$P(5\hat i + 7\hat j - 2\hat k)$$ and  $$Q( - 3\hat i + 3\hat j + 6\hat k)$$ parallel to $$(3\hat i - \hat j + \hat k)$$ and $$( - 3\hat i + 2\hat j + 4\hat k)$$ respectively. A line L parallel to $$(2\hat i + 7\hat j - 5\hat k)$$ intersects Land L2 in A and B

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