Question

Find the symmetrical form, the equation of the line formed by the plane $x + y + z + 1 = 0$ , $4x + y - 2z + 2 = 0$ and find its direction-cosines.
A. $\dfrac{{x - \dfrac{1}{3}}}{1} = \dfrac{{y + \dfrac{2}{3}}}{{ - 2}} = \dfrac{{z - 0}}{1}; - \dfrac{1}{{\sqrt 6 }},\dfrac{2}{{\sqrt 6 }},\dfrac{1}{{\sqrt 6 }}$
B. $\dfrac{{x - \dfrac{1}{3}}}{{ - 1}} = \dfrac{{y - \dfrac{2}{3}}}{2} = \dfrac{{z - 0}}{1};\dfrac{1}{{\sqrt 6 }},\dfrac{2}{{\sqrt 6 }}, - \dfrac{1}{{\sqrt 6 }}$
C. $\dfrac{{x + \dfrac{1}{3}}}{{ - 1}} = \dfrac{{y + \dfrac{2}{3}}}{2} = \dfrac{{z + 0}}{{ - 1}}; - \dfrac{1}{{\sqrt 6 }},\dfrac{2}{{\sqrt 6 }}, - \dfrac{1}{{\sqrt 6 }}$
D. $\dfrac{{x + \dfrac{1}{3}}}{1} = \dfrac{{y - \dfrac{2}{3}}}{2} = \dfrac{{z + 0}}{1};\dfrac{1}{{\sqrt 6 }}, - \dfrac{2}{{\sqrt 6 }},\dfrac{1}{{\sqrt 6 }}$

Answer 1

Hint : We know that equation of a line passing through the point $({x_0},{y_0},{z_0})$ and $(a,b,c)$ is $\dfrac{{x - {x_0}}}{a} = \dfrac{{y - {y_0}}}{b} = \dfrac{{z - {z_0}}}{c}$ .
Here we need to find the direction ratios and as well as coordinate points. We know that if two lines are perpendicular then direction ratios of two lines have dot product equal to zero.

Complete step by step solution:
We have $x + y + z + 1 = 0$ . This is the equation of the plane in Cartesian form with direction ratios $(1,1,1)$ .
Also we have $4x + y - 2z + 2 = 0$ This is the equation of the plane in Cartesian form with direction ratios $(4,1, - 2)$ .
Now, we know that the equation of line should lie on both planes. So the equation of the required line should be perpendicular to both the planes.
Let assume that the direction ratios of the required line is $(a,b,c)$ and we need to find this.
Since the direction ratios are perpendicular, that is
$(a,b,c) \bot (1,1,1)$
$(a,b,c) \bot (4,1, - 2)$
We know that if direction ratios of two planes are perpendicular then the dot product is zero.
Now for $(a,b,c) \bot (1,1,1)$ we have,
$(a \times 1) + (b \times 1) + (c \times 1) = 0$
$a + b + c = 0{\text{ }} - - - (1)$
Now for $(a,b,c) \bot (4,1, - 2)$ we have
$(a \times 4) + (b \times 1) + (c \times - 2) = 0$
$4a + b - 2c = 0{\text{ }} - - - - (2)$
Now solving (1) and (2) using cross product method that is,
If we have two linear equation with three variable, ${a_1}a + {b_1}b + {c_1}c = 0$ and ${a_2}a + {b_2}b + {c_2}c = 0$ then
$\dfrac{a}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{b}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{c}{{{a_1}{b_2} - {a_2}{b_1}}}$
Where ${a_1} = {b_1} = {c_1} = 1$ and ${a_2} = 4,{b_2} = 1,{c_2} = - 2$ . Substituting we have,
$\dfrac{a}{{(1 \times - 2) - (1 \times 1)}} = \dfrac{b}{{(1 \times 4) - ( - 2 \times 1)}} = \dfrac{c}{{(1 \times 1) - (4 \times 1)}}$
$\dfrac{a}{{ - 2 - 1}} = \dfrac{b}{{4 + 2}} = \dfrac{c}{{1 - 4}}$
$\dfrac{a}{{ - 3}} = \dfrac{b}{6} = \dfrac{c}{{ - 3}}$
Thus we have $a = - 3,b = 6$ and $c = - 3$
Or
Thus we have $a = - 1,b = 2$ and $c = - 1$
Thus the direction ratios are $( - 1,2, - 1)$
We have direction ratios now we need to find the coordinates.
Let the coordinates of the point lying in the xy-plane is $(x,y,0)$
Then equation $x + y + z + 1 = 0$ and $4x + y - 2z + 2 = 0$ becomes,
$x + y = - 1{\text{ }} - - - (3)$
$4x + y = - 2{\text{ }} - - - (4)$
Subtracting (3) and (4) we have,

$$\left( {x + y} \right) - \left( {4x + y} \right) = - 1 - ( - 2) \\\ x + y - 4x - y = - 1 + 2 \;$$

The term ‘y’ get cancels out we have,

$$x + - 4x = - 1 + 2 \\\ \- 3x = 1 \\\ x = - \dfrac{1}{3} \;$$

Now to find ‘y’ we substitute the value of ‘x’ in equation (3),

$$\- \dfrac{1}{3} + y = - 1 \\\ y = - 1 + \dfrac{1}{3} \\\ y = \dfrac{{ - 3 + 1}}{3} \\\ y = \dfrac{{ - 2}}{3} \;$$

Thus we have the coordinates $({x_0},{y_0},{z_0}) = \left( { - \dfrac{1}{3}, - \dfrac{2}{3},0} \right)$ and direction ratios are $(a,b,c) = ( - 1,2, - 1)$ ,
The equation is given by
$\dfrac{{x + \dfrac{1}{3}}}{{ - 1}} = \dfrac{{y + \dfrac{2}{3}}}{2} = \dfrac{{z + 0}}{{ - 1}}$
Now the direction cosine are given by,

$$DC = \left( {\dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right) \\\ = \left( {\dfrac{{ - 1}}{{\sqrt {{{\left( { - 1} \right)}^2} + {{(2)}^2} + {{( - 1)}^2}} }},\dfrac{2}{{\sqrt {{{\left( { - 1} \right)}^2} + {{(2)}^2} + {{( - 1)}^2}} }},\dfrac{{ - 1}}{{\sqrt {{{\left( { - 1} \right)}^2} + {{(2)}^2} + {{( - 1)}^2}} }}} \right) \\\ = \left( {\dfrac{{ - 1}}{{\sqrt {1 + 4 + 1} }},\dfrac{2}{{\sqrt {1 + 4 + 1} }},\dfrac{{ - 1}}{{\sqrt {1 + 4 + 1} }}} \right) \\\ = \left( {\dfrac{{ - 1}}{{\sqrt 6 }},\dfrac{2}{{\sqrt 6 }},\dfrac{{ - 1}}{{\sqrt 6 }}} \right) \;$$

We can see the required option is (C).
So, the correct answer is “Option C”.

Note : Since in above from equation (1) and (2) we have three variables with only two equations. So the cancelling method will be difficult and we don’t get the answer in that method so we use cross product methods which are easy to solve. Remember the cross product formula. Careful in the substitution and calculation part.