Question

if line x-1/2=y+1/3=z/4 lies in the plane 4x+4y-kz=0 the value of R is​

Answer 1

Step-by-step explanation:

There is not R there is K to find the value

Answer 2

Correct Statement :-

if line x-1/2=y+1/3=z/4 lies in the plane 4x+4y-kz=0 the value of k is

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

We know that

The general equation of line in cartesian form is given by

$\rm :\longmapsto\:\dfrac{x - p}{a} = \dfrac{y - q}{b} = \dfrac{z - r}{c}$

where,

• (p, q, r) is the point from where line passes.

• (a, b, c) is direction ratios of the line.

So,

• Given equation of line is

$\rm :\longmapsto\:\dfrac{x - 1}{2} = \dfrac{y + 1}{3} = \dfrac{z}{4}$

$\rm :\implies\:direction \: ratio \: of \: line \: is \: (2,3,4)$

We know direction ratios of line in vector form is represented as

$\rm :\longmapsto\: \vec{b} \: = \: 2 \hat{i} \: + 3 \: \hat{j} \: + \: 4 \hat{k} - - - (1)$

Now,

We know that,

• General equation of plane is given by

$\rm :\longmapsto\:ax + by + cz + d = 0$

where,

• (a, b, c) is normal with direction ratios to the plane

So,

• Given equation of plane is

$\rm :\longmapsto\:4x + 4y - kz = 0$

$\rm :\implies\:normal \: to \: pane \: with \: direction \: ratio\: is \: (4,4, - k)$

We know, the normal vector to the surface of plane is represented as

$\rm :\longmapsto\: \vec{n} \: = \: 4 \hat{i} \: + 4 \: \hat{j} \: - \: k \hat{k} - - - (2)$

Since, Line lies in the plane,

$\rm :\implies\: \vec{b}. \: \vec{n} = 0$

$\rm :\longmapsto\: \: (2 \hat{i} \: + 3 \: \hat{j} \: + \: 4 \hat{k}).( \: 4 \hat{i} \: + 4 \: \hat{j} \: - \: k \hat{k}) = 0$

$\rm :\longmapsto\:8 + 12 - 4k = 0$

$\rm :\longmapsto\:20 - 4k = 0$

$\rm :\implies\:k \: = \: 5$

Additional Information :-

☆ Let us consider two lines,

$\rm :\longmapsto\:\vec{r} = \vec{a_1} + \alpha \vec{b_1} \: \: and \: \vec{r} = \vec{a_2} + \alpha \vec{b_2}$

1. Shortest distance between two non parallel lines is

$\: \sf \: \: = \: \: \bigg|\dfrac{(\vec{a_2} - \vec{a_1}).(\vec{b_1 \times \vec{b_2)}}}{ |\vec{b_1} \times \vec{b_2}| } \bigg|$

2. Angle between lines is given by

$\rm :\longmapsto\:cos \theta \: = \dfrac{\vec{b_1}.\vec{b_2}}{ |\vec{b_1}| |\vec{b_2}| }$

3. Condition for perpendicular lines :-

$\rm :\longmapsto\:Lines \: are \perp \: iff \: \vec{b_1}.\vec{b_2} = 0$

☆ Let us consider two parallel lines,

$\rm :\longmapsto\:\vec{r} = \vec{a_1} + \alpha \vec{b} \: \: and \: \: \vec{r} = \vec{a_2} + \alpha \vec{b}$

Shortest distance between two parallel lines is

$\: \sf \: \: = \: \: \bigg|\dfrac{(\vec{a_2} - \vec{a_1}) \times \vec{b}}{ |\vec{b}| } \bigg|$