Question

The direction cosines of the normal to the plane $x + 2y - 3z + 4 = 0$ are:
A. $\dfrac{{ - 1}}{{\sqrt {14} }},\dfrac{{ - 2}}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}$
B. $\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}$
C. $\dfrac{{ - 1}}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}$
D. $\dfrac{1}{{\sqrt {14} }},\dfrac{{ - 2}}{{\sqrt {14} }},\dfrac{{ - 3}}{{\sqrt {14} }}$

Answer 1

Hint: We are given an equation of the plane and we are supposed to find the directional cosines. To solve this we are going to use the formula for directional cosine and equating the given equation to find the vector coordinates which would further be used in the formula.

Formula Used:
Directional cosines of normal to a plane $ax + by + cz + d = 0$is:
\[ \pm \left( {\dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right)\] \[ - - - - - - eq(1)\]

Complete Step-by-step solution
As we have the equation (1) for finding the directional cosines of normal to a plane, we can equate the given equation $x + 2y - 3z + 4 = 0$ with equation (1)
On equating we get:
$a = 1,b = 2,c = - 3,d = 4$
Now, on putting the values in the equation (1) we get:
Directional cosines \[ = \pm \left( {\dfrac{1}{{\sqrt {{1^2} + {2^2} + {{( - 3)}^2}} }},\dfrac{2}{{\sqrt {{1^2} + {2^2} + {{( - 3)}^2}} }},\dfrac{{ - 3}}{{\sqrt {{1^2} + {2^2} + {{( - 3)}^2}} }}} \right)\]
\[ = \pm \left( {\dfrac{1}{{\sqrt {1 + 4 + 9} }},\dfrac{2}{{\sqrt {1 + 4 + 9} }},\dfrac{{ - 3}}{{\sqrt {1 + 4 + 9} }}} \right)\]
\[ = \pm \left( {\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{{ - 3}}{{\sqrt {14} }}} \right)\]
So, the required directional cosines are \[ + \left( {\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{{ - 3}}{{\sqrt {14} }}} \right)\]and\[ - \left( {\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{{ - 3}}{{\sqrt {14} }}} \right)\]

It can be written as \[ + \left( {\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{{ - 3}}{{\sqrt {14} }}} \right)\]and \[\left( {\dfrac{{ - 1}}{{\sqrt {14} }},\dfrac{{ - 2}}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}} \right)\]

Hence the correct option is A.

Note:If the plane equation is given in the question, we can easily find the directional cosines of normal to that plane using the formula or the equation. Also, keep an eye on the positive and negative signs.