Question

The area of the figure formed by the points (-1,-1,1);(1,1,1) and the mirror images with respect to the plane 3x+2y+4z+1=0

Answer 1

Given :

Points of quadrilateral = (-1,-1,1) and (1,1,1)

Equation of plane = 3x + 2y + 4z +1 = 0

To find :

Area of figure formed by points and their image with respect to given plane.

Solution :

To get the area of this figure we firstly have to find the other two points which are the mirror images of given two points with respect to given plane . after finding these two points we can find the area of the figure.

Formula to find the mirror points with respect to given plane ( ax + by + cz + d = 0) :

$\\ \bf\frac{(x - x_0)}{a} = \frac{(y - y_0)}{b} = \frac{(z - z_0)}{c} = \frac{ - 2(ax_0 + by_0 + cz_0 + d)}{ {a}^{2} + {b}^{2} + {c}^{2} } \:$

(equation 1)

For points A (-1,-1,1) and plane 3x + 2y + 4z +1 = 0

here (a,b,c,d) = (3,2,4,1)

The mirror points will be :

$\\ \bf\frac{(x + 1)}{3} = \frac{(y + 1)}{2} = \frac{(z - 1)}{4} = \frac{ - 2((3 \times -1) + (2 \times - 1) + (4 \times 1) + 1)}{ {3}^{2} + {2}^{2} + {4}^{2} } \: = \frac{2}{29}$

on solving this, we get point C :

$\\ \bf \: x_0 = - \frac{23}{29} \\ \bf \: y_0 = - \frac{25}{29} \\ \bf \: z_0 = \frac{37}{29}$

Similarly for points B (1,1,1) on plane 3x + 2y + 4z +1 = 0

Putting in equation 1, we get

$\\ \bf\frac{(x - 1)}{3} = \frac{(y - 1)}{2} = \frac{(z - 1)}{4} = \frac{ - 2((3 \times -1) + (2 \times - 1) + (4 \times 1) + 1)}{ {3}^{2} + {2}^{2} + {4}^{2} } \: = \frac{2}{29} \:$

on solving this, we get point D :

$\\ \bf \: x_0 = \frac{35}{29} \\ \bf \: y_0 = \frac{33}{29} \\ \bf \: z_0 = \frac{37}{29}$

To find the area of quadrilateral,

we will choose two sides adjacent to each other,

so we will find AB as

$\bf \: A = ( -1, -1, 1), B = (1,1,1) \\ </strong></p><p><strong>[tex] \bf \: A = ( -1, -1, 1), B = (1,1,1) \\ \bf \: So \\ \bf \: AB = B - A = (2,2,0)$

and AC as,

$\bf \: A = (-1, -1, 1) \\ </p><p> \bf \: C = (\frac{-23}{ 29} , \frac{-25 }{29}, \frac { 37 }{29}) \: \\ \bf so \\ \bf \: AC = C - A \: = (\frac{6}{ 29} , \frac{4 }{29}, \frac {8 }{29}) \:$

so,

Area of quadrilateral :

$\bf | | AB \times AC | |$

(equation 2)

So, Area of quadrilateral is the magnitude of cross product of two adjacent sides of quadrilateral

By putting the value of AB and AC in equation 2,

$</p><p> \bf cross \: product = | AB \times AC | = (2,2,0) \times (\frac{6}{ 29} , \frac{4}{29}, \frac {8}{29}) \\ \bf = \frac{4}{29}( (1,1,0) \times (3,2,4)) \\ \bf = \frac{4}{29} (\hat i(4 - 0) - \hat j(4 - 0) + \hat k(2 - 3))$

So,

Cross product of AB and AC is :

$\bf = \frac{4}{29} (4 \hat i - 4 \hat j \: - \hat k)$

So,

Area of quadrilateral is the magnitude of this cross product :

$\bf \: Area = \frac{4}{29} \sqrt{ {4}^{2} + {4}^{2} + {1}^{2} } \\ \bf \: Area = \frac{4}{29} \sqrt{33}$