Question

Find the vector and Cartesian equation of the plane passing

through the points A(2 ,3 , 4 ) B ( 4 , -5 , ,3 ) and parallel to X axis.


PLS ANSWER FAST

Answer 1

$\large\underline{\bold{Solution-}}$

We know that,

The general equation of plane which passes through the point (u, v, w) is given by

• a(x -- u) + b(y - v) + c(z - w) = 0,

where a, b, c are real constant numbers.

So,

The required equation of plane which passes through (2, 3, 4) is given by

$\sf \: a(x - 2) + b(y - 3) + c(z - 4) = 0 - - (1)$

Since, (1) passes through (4, - 5, 3), then

$\rm :\longmapsto\:\sf \: a(4 - 2) + b( - 5 - 3) + c(3 - 4) = 0$

$\rm :\longmapsto\:2a - 8b - c = 0 - - (2)$

We know,

Equation of x - axis is

$\rm :\longmapsto\:\dfrac{x}{1} = \dfrac{y}{0} = \dfrac{z}{0}$

So,

$\sf \: direction \: ratio \: of \: x \: - \: axis \: is \: (1, \: 0 , \: 0)$

Also,

• Direction ratios of plane (1) is (a, b, c).

Since, plane (1) is parallel to x - axis.

Therefore,

$\rm :\longmapsto\:a.1 \: + b.0 \: + c.0 = 0$

$\bf\implies \:a \: = \: 0 - - - (3)$

On substituting the value of 'a' in equation (1), then

$\rm :\longmapsto\:2 \times 0 - 8b - c = 0$

$\rm :\longmapsto\:8b = - \: c$

$\bf\implies \:c \: = \: - \: 8b - - (4)$

On Substituting the value of 'a' and 'c' in equation (1), we get

$\sf \: 0 \times (x - 2) + b(y - 3) - 8b(z - 4) = 0$

$\rm :\longmapsto\:b\bigg( y - 3 - 8z + 32 \bigg) = 0$

$\bf\implies \:y \: - \: 8z = - 29 \: is \: required \: equation \: of \: plane.$

Thus,

In vector form,

The equation of plane y - 8z = - 29 can be represented by

$\rm :\longmapsto\: \vec{r}. \: ( \: \hat{j} \: - \: 8 \: \hat{k} \: ) \: = \: - \: 29$

Additional Information :-

1. Angle between two planes :-

$\rm :\longmapsto\:cos \theta \: = \: \bigg |\dfrac{\vec{n_{1}}.\vec{n_{2}}}{ |\vec{n_{1}}| |\vec{n_{2}}| } \bigg|$

$\sf \: where$

$\: \: \: \: \: \: \: \bull \: \sf \vec{n_{1}} \: is \: normal \: vector \: of \: plane \: 1$

$\: \: \: \: \: \: \: \bull \: \sf \vec{n_{2}} \: is \: normal \: vector \: of \: plane \: 2$

$2. \: \: \sf \: 2 \: planes \: are \: parallel \: iff \: \: \vec{n_{1}} \: = \: k \: \vec{n_{2}}$

$3. \: \: \sf \: 2 \: planes \: are \: perpendicular \: iff \: \vec{n_{1}}.\vec{n_{2}} = 0.$

Answer 2

Answer:

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