Mathematics : Applications of Vector Algebra: Application of Vectors to 3-Dimensional Geometry

Theorem 6.12

The parametric form of vector equation of a line passing through two given points whose position vectors are * *and * *respectively is , *t *âˆˆ R.

(b) Non-parametric form of vector equation

The above equation can be written equivalently in non-parametric form of vector equation as

=

Suppose *P *is (*x*, *y*, *z*) , *A *is (*x*1, *y*1 , *z*1 ) and *B *is (*x*2 , *y*2 , *z*2). Then substituting * *= *x* *Ë†i *+ *y* *Ë† j *+ *z* *Ë†k *, * *= *x*1*Ë†i *+ *y*1*Ë†* *j *+ *z*1*Ë†k and ** *= *x*2*i *+ *y*2 *Ë†* *j *+*z*2*Ë†k *in theorem 6.12 and comparing the coefficients of *Ë†i *,* Ë†* *j*, *Ë†k *, we get *x *âˆ’ *x*1 = *t*(*x*2 âˆ’ *x*1), *y *âˆ’ *y*1 = *t*( *y*2 âˆ’ *y*1), *z *âˆ’ *z*1 = *t*(*z*2 âˆ’ *z*1 ) and so the Cartesian equations of a line passing through two given points (*x*1, *y*1, *z*1) and (*x*2, *y*2, *z*2) are given by

From the above equation, we observe that the direction ratios of a line passing through two given points (*x*1 , *y*1 , *z*1) and (*x*2 , *y*2 , *z*2 ) are given by* x*2âˆ’ *x*1 , *y*2 âˆ’ *y*1 , *z*2 âˆ’ *z*1, which are also given by any three numbers proportional to them and in particular* x*1 âˆ’ *x*2 , *y*1 âˆ’ *y*2 , *z*1 âˆ’ *z*2.

Example 6.24

A straight line passes through the point (1, 2, âˆ’3) and parallel to 4*i*Ë† + 5 Ë†*j *âˆ’ 7*k*Ë† . Find (i) vector equation in parametric form (ii) vector equation in non-parametric form (iii) Cartesian equations of the straight line.

*Solution*

The required line passes through (1, 2, âˆ’3) . So, the position vector of the point is *i*Ë† + 2 Ë†*j *âˆ’ 3*k*Ë†.

Let * *= *Ë†**i *+ 2 *Ë†**j *âˆ’ 3*Ë†**k *and * *= 4*Ë†**i *+ 5 *Ë†**j *âˆ’ 7*Ë†**k *. Then, we have

Let a = i + 2 j - 3k and b = 4i + 5 j - 7k . Then, we have

(i) vector equation of the required straight line in parametric form is = * + t**, t *âˆˆ R.

Therefore, = (Ë†*i + 2* Ë†* j - 3* Ë†*k ) + t(4* Ë†*i + 5* Ë†* j - 7* Ë†* k ), t*âˆˆ R.*.*

(ii) vector equation of the required straight line in non-parametric form is *( ** - **) Ã— * = .

Therefore, ( - (Ë†*i + 2* Ë†*j - 3* Ë†*k *)) *Ã—* (4 Ë†*i + 5* Ë†*j - 7* Ë†*k* ) = .

(iii) Cartesian equations of the required line are *(x - x1) / b1 = y - y1 / b1 = (z - z1) / b1.*

Here, (*x*1 , *y*1 , *z*1) = (1, 2, -3) and direction ratios of the required line are proportional to 4, 5, -7 . Therefore, Cartesian equations of the straight line are (*x* -1)/4 = (*y* â€“ 2)/5 = (*z* + 3)/-7.

Example 6.25

The vector equation in parametric form of a line is * *= (3 Ë†*i *âˆ’ 2 Ë†*j *+ 6 Ë†*k *) + *t*(2 Ë†*i *âˆ’ Ë†*j *+ 3 Ë†*k *) . Find (i) the direction cosines of the straight line (ii) vector equation in non-parametric form of the line (iii)Cartesian equations of the line.

Solution

Comparing the given equation with equation of a straight line * *= * *+ *t** *, we have = 3 Ë†*i *âˆ’ 2 Ë†*j *+ 6 Ë†*k *and * *= 2*i*Ë† âˆ’ Ë†*j *+ 3*k*Ë† . Therefore,

(i) If = b1*i*Ë† + b2Ë†*j* + b3*k*Ë† , then direction ratios of the straight line are b1 , b2 , b3. Therefore, direction ratios of the given straight line are proportional to 2, -1, 3 , and hence the direction cosines of the given straight line are .

(ii) vector equation of the straight line in non-parametric form is given by ( - ) Ã— = . Therefore, ( - (3 Ë†*i* - 2 Ë†*j* + 6 Ë†*k* )) x(2 Ë†*i* - Ë†*j* + 3 Ë†*k*) = 0 .

(iii) Here (x1 , y1 , z1 ) = (3, -2, 6) and the direction ratios are proportional to 2, -1, 3 .

Therefore, Cartesian equations of the straight line are (x â€“ 3)/2 = (y + 2)/-1 = (z â€“ 6)/3

Example 6.26

Find the vector equation in parametric form and Cartesian equations of the line passing through (âˆ’4, 2, âˆ’3) and is parallel to the line

Solution

Rewriting the given equations as and comparing with

We have

Clearly, is parallel to the vector 8*i*Ë† + 4Ë†*j* - 3*k*Ë† . Therefore, a vector equation of the required straight line passing through the given point (-4, 2, -3) and parallel to the vector 8*i*Ë† + 4Ë†*j* - 3*k*Ë† in parametric form is

= (-4*i*Ë† + 2Ë†*j* - 3*k*Ë†) + t(8*i*Ë† + 4Ë†*j* - 3*k*Ë†), t âˆˆ R.

Therefore, Cartesian equations of the required straight line are given by

(x + 4) / 8 = (y â€“ 2) / 4 = (z + 3) / -3 .

Find the vector equation in parametric form and Cartesian equations of a straight passing through the points (âˆ’5, 7, âˆ’4) and (13, âˆ’5, 2) . Find the point where the straight line crosses the *xy *-plane.

The straight line passes through the points (âˆ’5, 7, âˆ’4) and (13, âˆ’5, 2) , and therefore, direction ratios of the straight line joining these two points are 18, âˆ’12, 6 . That is 3, âˆ’2,1.

So, the straight line is parallel to 3*i*Ë† âˆ’ 2 Ë†*j *+ *k*Ë† . Therefore,

Required vector equation of the straight line in parametric form is * *= (âˆ’5Ë†*i *+ 7 Ë†*j *âˆ’ 4Ë†*k *) + *t*(3Ë†*i *âˆ’ 2Ë†*j *+ Ë†*k *) or* ** *= (13Ë†*i *âˆ’ 5Ë†*j *+ 2Ë†*k *) + *s*(3Ë†*i *âˆ’ 2Ë†*j *+Ë†*k *) where *s*, *t *âˆˆ R.

Required cartesian equations of the straight line are

An arbitrary point on the straight line is of the form

Since the straight line crosses the xy -plane, the z -coordinate of the point of intersection is zero. Therefore, we have t âˆ’ = 4 0 , that is, t = 4, and hence the straight line crosses the xy -plane at (7,âˆ’1,0).

Find the angles between the straight line with coordinate axes.

If *b*Ë† is a unit vector parallel to the given line, then *b*Ë† = Therefore, from the definition of direction cosines of *b*Ë† , we have

where Î± , Î² ,Î³ are the angles made by *b*Ë† with the positive x -axis, positive y -axis, and positive z -axis, respectively. As the angle between the given straight line with the coordinate axes are same as the angles made by *b*Ë† with the coordinate axes, we have Î± = cos-1 (2/3), Î² = cos-1( 2/3), Î³ = cos-1(-1/3), respectively.

Tags : Definition, Theorem, Proof, Solved Example Problems, Solution , 12th Mathematics : UNIT 6 : Applications of Vector Algebra

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

12th Mathematics : UNIT 6 : Applications of Vector Algebra : Straight Line passing through two given points | Definition, Theorem, Proof, Solved Example Problems, Solution

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright Â© 2018-2024 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.