In the previous section, we completed a discussion related to the intersection of two planes. In this section, we will see coplanarity of two lines.
The vector form can be obtained in 7 steps:
1. In fig.27.24 below, two magenta lines lie on the same plane.
• That is., the two lines are coplanar lines.
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| Fig.27.24 |
2. We are given the following information about the two lines:
(i) The first line passes through the point $\small{U_1}$.
♦ The position vector of $\small{U_1~\text{is}~\vec{u_1}}$
♦ This line is parallel to $\small{\vec{v_1}}$
(ii) The second line passes through the point $\small{U_2}$.
♦ The position vector of $\small{U_2~\text{is}~\vec{u_2}}$
♦ This line is parallel to $\small{\vec{v_2}}$
3. Now we write the vectors which lie flat on the plane:
(i) First line is parallel to $\small{\vec{v_1}}$. So this $\small{\vec{v_1}}$ can be shifted and made to lie flat on the plane.
(ii) The second line is parallel to $\small{\vec{v_2}}$. So this $\small{\vec{v_2}}$ can also be shifted and made to lie flat on the plane.
4. Since $\small{\vec{v_1}~\text{and}~\vec{v_2}}$ lie on the plane, the cross product $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$ will be a vector perpendicular to the plane.
5. Based on the position vectors of the points $\small{U_1~\text{and}~U_2}$, we get:
$\small{\vec{U_1 U_2}=\vec{u_2}-\vec{u_1}}$
6. Now we see an interesting situation:
• $\small{\vec{U_1 U_2}}$ lies on the plane, while $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$ is perpendicular to the plane.
• So their dot product must be zero.
7. Therefore we can write:
• The given lines are coplanar if and only if $\small{\vec{U_1 U_2}.\left(\vec{v_1}\times\vec{v_2} \right)=0}$
• In other words, the given lines are coplanar if and only if $\small{\left(\vec{u_2}-\vec{u_1} \right).\left(\vec{v_1}\times\vec{v_2} \right)=0}$
• This is the vector form of the condition
The Cartesian form can be obtained in steps:
1. First we write the coordinates:
♦ The coordinates of $\small{U_1~\text{are}~\left(x_1,y_1,z_1 \right)}$
♦ The coordinates of $\small{U_2~\text{are}~\left(x_2,y_2,z_2 \right)}$
2. Based on the above coordinates, we can write $\small{\vec{U_1 U_2}}$ in component form:
$\small{\vec{U_1 U_2} = \left(x_2 - x_1 \right)\hat{i} + \left(y_2 - y_1 \right)\hat{j} + \left(z_2 - z_1 \right)\hat{k}}$
3. Next we write the direction ratios of $\small{\vec{v_1}~\text{and}~\vec{v_2}}$:
♦ Direction ratios of $\small{\vec{v_1}~\text{are}~a_1,~b_1,~c_1}$
♦ Direction ratios of $\small{\vec{v_2}~\text{are}~a_2,~b_2,~c_2}$
4. Based on the above direction ratios, we can write $\small{\vec{v_1}~\text{and}~\vec{v_2}}$ in component forms:
♦ $\small{\vec{v_1}=a_1\hat{i}+b_1\hat{j}+c_1\hat{k}}$
♦ $\small{\vec{v_2}=a_2\hat{i}+b_2\hat{j}+c_2\hat{k}}$
5. We obtained the condition in vector form as:
$\small{\left(\vec{u_2}-\vec{u_1} \right).\left(\vec{v_1}\times\vec{v_2} \right)=0}$
6. First we will write $\small{\left(\vec{u_2}-\vec{u_1} \right)}$
We have:
$\small{\left(\vec{u_2}-\vec{u_1} \right)=\vec{U_1 U_2} = \left(x_2 - x_1 \right)\hat{i} + \left(y_2 - y_1 \right)\hat{j} + \left(z_2 - z_1 \right)\hat{k}}$
7. Next we write $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$
We have:
$\vec{v_1}\times\vec{v_2}~=~\left|\begin{array}{r} \hat{i} &{ \hat{j} } &{ \hat{k} } \\ a_{1} &{ a_{2} } &{ a_{3} } \\ b_{1} &{ b_{2} } &{ b_{3} } \\ \end{array}\right|$
• Expanding this, we get:
$\small{\vec{v_1}\times\vec{v_2}=\left(a_2 b_3 - a_3 b_2 \right)\hat{i}-\left(a_1 b_3 - a_3 b_1 \right)\hat{j}+\left(a_1 b_2 - a_2 b_1 \right)\hat{k}}$
8. So the vector condition written in (5) is:
$\small{\left(x_2 - x_1 \right)\left(a_2 b_3 - a_3 b_2 \right)\hat{i}-\left(y_2 - y_1 \right)\left(a_1 b_3 - a_3 b_1 \right)\hat{j}+\left(z_2 - z_1 \right)\left(a_1 b_2 - a_2 b_1 \right)\hat{k}=0}$
9. By close examination, we see that, the L.H.S in the above result is a determinant. We can write:
$\left|\begin{array}{r} x_2 - x_1 &{ y_2 - y_1 } &{ z_2 - z_1 } \\ a_{1} &{ a_{2} } &{ a_{3} } \\ b_{1} &{ b_{2} } &{ b_{3} } \\ \end{array}\right|~=~0$
• This is the Cartesian form of the condition
Now we will see a solved example
Solved example 27.53
Show that the lines
$\small{\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}~\text{and}~\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}}$ are coplanar.
Solution:
Part (a): Showing coplanarity in vector form
1. The given lines in vector form are:
• $\small{-3\hat{i}+\hat{j}+5\hat{k}+\lambda \left(-3\hat{i}+\hat{j}+5\hat{k} \right)}$
• $\small{-\hat{i}+2\hat{j}+5\hat{k}+\lambda \left(-\hat{i}+2\hat{j}+5\hat{k} \right)}$
2. Form the above vector forms, we get:
• $\small{\vec{u_1}= -3\hat{i}+\hat{j}+5\hat{k}}$
• $\small{\vec{u_2}= -\hat{i}+2\hat{j}+5\hat{k}}$
• $\small{\vec{v_1}= -3\hat{i}+\hat{j}+5\hat{k}}$
• $\small{\vec{v_2}= -\hat{i}+2\hat{j}+5\hat{k}}$
3. $\small{\vec{u_2}-\vec{u_1}=2\hat{i}+\hat{j}}$
4. $\small{\vec{v_1}\times\vec{v_2}=-5\hat{i}+10\hat{j}-5\hat{k}}$
(The reader should write all the steps related to the above cross product)
5. The given lines are coplanar if and only if the dot product of the vectors in (3) and (4) is zero.
• The dot product is:
(2)(−5)+(1)(10)+(0)(-5) = −10 + 10 = 0
• So the lines are coplanar.
Part (b): Showing coplanarity in Cartesian form
1. From the given equations, we get:
• $\small{\left(x_1,y_1,z_1 \right)=\left(-3,1,5 \right)}$
• $\small{\left(x_2,y_2,z_2 \right)=\left(-1,2,5 \right)}$
• $\small{a_1,b_1,c_1~=~-3,1,5}$
• $\small{a_2,b_2,c_2~=~-1,2,5}$
2. The condition in Cartesian form is:
$\left|\begin{array}{r} x_2 - x_1
&{ y_2 - y_1 } &{ z_2 - z_1 } \\ a_{1}
&{ a_{2} } &{ a_{3} } \\ b_{1}
&{ b_{2} } &{ b_{3} } \\
\end{array}\right|~=~0$
3. Let us evaluate the determinant:
$\left|\begin{array}{r} (-1)-(-3)&{ 2-1 } &{ 5-5 } \\ -3 &{ 1 } &{ 5 } \\ -1 &{ 2 } &{ 5 } \\ \end{array}\right|$
$\small{~=(2)(5-10)-(1)(-15+5)+(0)(-6+1)}$
$\small{~=(2)(-5)-(1)(-10)}$
$\small{~=-10+10=0}$
• So the given lines are coplanar
In the next section, we will see angle between two planes.
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