Question

Find the vector equation of the plane which is at a distance of $5$units from the origin and normal to the plane is $3\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge $.

Answer 1

Hint: We have to find the vector equation of the plane using the formula $\mathop r\limits^ \to.\mathop x\limits^ \wedge = $ d.
It is given that the normal vector of the plane is $3\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge $. Using this we can find the value of $\mathop x\limits^ \wedge $, substituting the value of $\mathop x\limits^ \wedge $ and the value of distance (d) in the formula of vector equation of the plane, we can find the required vector equation of the plane.

Complete step-by-step solution:
Using the formula we are going to find the vector equation of the plane which is at the distance of $5$units, the vector equation of the plane is, $\mathop r\limits^ \to.\mathop x\limits^ \wedge = $ d, Where d is the distance from origin that is $5$units
Given that the normal vector of the plane is $3\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge $.
i.e., $\mathop x\limits^ \to = $$3\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge $.
We know that $\mathop x\limits^ \wedge $=$\dfrac{{\mathop x\limits^ \to }}{{\left| {\mathop x\limits^ \to } \right|}}$……….. (1)
The magnitude of the vector equation $\left| {\mathop x\limits^ \to} \right|$ = $\sqrt {{3^2} + {2^2} + {6^2}} $
Squaring all the terms inside the square root, we get
$\left| {\mathop x\limits^ \to} \right|$
= $\sqrt {9 + 4 + 36} $
= $\sqrt {49} $
= $7$
Substituting all the values in equation (1), we get
$\mathop x\limits^ \wedge $=$\dfrac{{}}{{}}$$\dfrac{{3\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge }}{7}$
Now we know the formula to find the vector equation of the plane, i.e., vector equation of the plane is, $\mathop r\limits^ \to .\mathop x\limits^ \wedge = $ d
Substituting the value of $\mathop x\limits^ \wedge $ in the above formula, we get
$\mathop r\limits^ \to $. $\dfrac{{3\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge }}{7}$ = d
Now substituting the value of distance = $5$ units in the formula, we get
$\mathop r\limits^ \to $. $\dfrac{{3\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge }}{7}$ = 5
                    (or)
Taking 7 to the left hand side, we get
$\mathop r\limits^ \to $($3\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge $) = $35$.
Therefore, the required vector equation of the plane is $\mathop r\limits^ \to .$($3\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge $) = $35$.

Note: To find the magnitude only scalar values must be taken not the vectors. For different distances and normals to the plane, the vector equation will be changing. If the distance from the origin and normal to the plane is not given, then we have to use a different formula to find the vector equation.