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Definition, Theorem, Proof, Solved Example Problems, Solution - Application of Vectors to 3-Dimensional Geometry | 12th Mathematics : UNIT 6 : Applications of Vector Algebra

Chapter: 12th Mathematics : UNIT 6 : Applications of Vector Algebra

Application of Vectors to 3-Dimensional Geometry

Vectors provide an elegant approach to study straight lines and planes in three dimension.

Application of Vectors to 3-Dimensional Geometry

Vectors provide an elegant approach to study straight lines and planes in three dimension. All straight lines and planes are subsets of R3. For brevity, we shall call a straight line simply as line. A plane is a surface which is understood as a set P of points in R3 such that , if A, B, and C are any three non-collinear points of P , then the line passing through any two of them is a subset of P. Two planes are said to be intersecting if they have at least one point in common and at least one point which lies on one plane but not on the other. Two planes are said to be coincident if they have exactly the same points. Two planes are said to be parallel but not coincident if they have no point in common. Similarly, a straight line can be understood as the set of points common to two intersecting planes. In this section, we obtain vector and Cartesian equations of straight line and plane by applying vector methods. By a vector form of equation of a geometrical object, we mean an equation which is satisfied by the position vector of every point of the object. The equation may be a vector equation or a scalar equation.

 

1. Different forms of equation of a straight line

A straight line can be uniquely fixed if

·            a point on the straight line and the direction of the straight line are given

·            two points on the straight line are given

We find equations of a straight line in vector and Cartesian form. To find the equation of a straight line in vector form, an arbitrary point P with position vector  on the straight line is taken and a relation satisfied by  is obtained by using the given conditions. This relation is called the vector equation of the straight line. A vector equation of a straight line may or may not involve parameters. If a vector equation involves parameters, then it is called a vector equation in parametric form. If no parameter is involved, then the equation is called a vector equation in non – parametric form.

 

2. A point on the straight line and the direction of the straight line  are given

(A) Parametric form of vector equation

Theorem 6.11

The vector equation of a straight line passing through a fixed point with position vector  and parallel to a given vector  is  =  + t, where t R.

Proof

If  is the position vector of a given point A and  is the position vector of an arbitrary point  P on the straight line, then

 .


This is the vector equation of the straight line in parametric form.

Remark

The position vector of any point on the line is taken as  + t.

(b) Non-parametric form of vector equation

Since  is parallel to  , we have   ×  

That is, ( − ) ×  = 0 .

This is known as the vector equation of the straight line in non-parametric form.

(c) Cartesian equation

Suppose P is (x, y, z) , A is (x1 , y1 , z1 ) and  = b1 ˆi + b2 ˆj + b3 ˆk . Then, substituting  = x ˆi + y ˆj + z ˆk ,  = x1ˆi + y1ˆ j + z1 ˆk  in (1) and comparing the coefficients of ˆi , ˆj, ˆk , we get

x x1 = tb1 , y y1 = tb2 , z z1 = tb3                     ………….(4)

Conventionally (4) can be written as


which are called the Cartesian equations or symmetric equations of a straight line passing through the point (x1, y1 , z1) and parallel to a vector with direction ratios b1, b2, b3.

Remark

(i) Every point on the line (5) is of the form (x1 + tb1 , y1 + tb2 , z1 + tb3) , where t R.

(ii) Since the direction cosines of a line are proportional to direction ratios of the line, if l, m, n are the direction cosines of the line, then the Cartesian equations of the line are


(iii) In (5), if any one or two of b1, b2 , b3 are zero, it does not mean that we are dividing by zero. But it means that the corresponding numerator is zero. For instance, If b1 ≠ 0, b2 ≠ 0 and b3 = 0 , then 


(iv) We know that the direction cosines of x - axis are 1, 0,0 . Therefore, the equations of x -axis are 


Similarly the equations of y -axis and z -axis are given by  respectively.

 

3. Straight Line passing through two given points

(a) Parametric form of vector equation

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 R.

(b) Non-parametric form of vector equation

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

  = 

(c) Cartesian form of equation

Suppose P is (x, y, z) ,  A is (x1, y1 , z1 ) and B is (x2 , y2 , z2). Then substituting  = x ˆi + y ˆ j + z ˆk ,  = x1ˆi + y1ˆ j + z1ˆk and  = x2i + y2 ˆ j + z2ˆk in theorem 6.12 and comparing the coefficients of ˆi , ˆ j, ˆk , we get  x x1  = t(x2x1), y y1  = t( y2y1), z z1  = t(z2z1 ) and so the Cartesian equations of a  line passing through two given points (x1, y1, z1) and (x2, y2, z2) are given by


From the above equation, we observe that the direction ratios of a line passing through two given points (x1 , y1 , z1) and (x2 , y2 , z2 ) are given by x2x1 , y2y1 , z2z1, which are also given by any three numbers proportional to them and in particular x1x2 , y1y2 , z1z2.

 

Example 6.24

A straight line passes through the point (1, 2, −3) and parallel to 4iˆ + 5 ˆj − 7kˆ . 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, −3) . So, the position vector of the point is iˆ + 2 ˆj − 3kˆ.

Let  = ˆi + 2 ˆj − 3ˆk and  = 4ˆi + 5 ˆj − 7ˆk . 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 R.

Therefore,  = (ˆi + 2 ˆ j - 3 ˆk ) + t(4 ˆi + 5 ˆ j - 7 ˆ k ), tR..

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

Therefore, (  - (ˆi + 2 ˆj - 3 ˆk )) × (4 ˆi + 5 ˆj - 7 ˆk ) = .

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

Here, (x1 , y1 , z1) = (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 – 2)/5 = (z + 3)/-7.

 

Example 6.25

The vector equation in parametric form of a line is  = (3 ˆi − 2 ˆj + 6 ˆk ) + t(2 ˆi − ˆj + 3 ˆk ) . 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 ˆi − 2 ˆj + 6 ˆk and  = 2iˆ − ˆj + 3kˆ . Therefore,

(i)   If  = b1iˆ + b2ˆj + b3kˆ ,  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  (  -  ) ×  =  . Therefore, (  - (3 ˆi - 2 ˆj + 6 ˆk )) x(2 ˆi - ˆj + 3 ˆk ) = 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 – 3)/2 = (y + 2)/-1 = (z – 6)/3

 

Example 6.26

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

Solution

Rewriting the given equations as  and comparing with 

We have 

Clearly,  is parallel to the vector 8iˆ + 4ˆj - 3kˆ . Therefore, a vector equation of the required straight line passing through the given point (-4, 2, -3) and parallel to the vector 8iˆ + 4ˆj - 3kˆ in parametric form is

   = (-4iˆ + 2ˆj  - 3kˆ) + t(8iˆ + 4ˆj - 3kˆ), t R.

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

(x + 4) / 8 =  (y – 2) / 4 = (z + 3) / -3 .

 

Example 6.27

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

Solution

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

So, the straight line is parallel to 3iˆ − 2 ˆj + kˆ . Therefore,

Required vector equation of the straight line in parametric form is  = (−5ˆi + 7 ˆj − 4ˆk ) + t(3ˆi − 2ˆj + ˆk ) or  = (13ˆi − 5ˆj + 2ˆk ) + s(3ˆi − 2ˆj + ˆk ) where s, t 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 − = 4 0 , that is, t = 4, and hence the straight line crosses the xy -plane at (7,−1,0).

 

Example 6.28

Find the angles between the straight line  with coordinate axes.

Solution

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


where α , β ,γ are the angles made by bˆ 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 bˆ with the coordinate axes, we have α = cos-1 (2/3), β = cos-1( 2/3), γ = cos-1(-1/3), respectively.

 

4. Angle between two straight lines

(a) Vector form

The acute angle between two given straight lines


(b) Cartesian form

If two lines are given in Cartesian form as  then the acute angle θ between the two given lines is given by


Remark

(i) The two given lines with direction ratios b1 , b2 , b3 and d1 , d2 , d3 are parallel if, and only if .

(ii)  The two given lines with direction ratios b1, b2, b3 and d1, d2, d3 are perpendicular if and only if b1d1 + b2d2 + b3d3 = 0 .

(iii) If the direction cosines of two given straight lines are l1 , m1 , n1 and l2 , m2 , n2, then the angle between the two given straight lines is cos θ =| l1l2 + m1m2 + n1n2 | .

 

Example 6.29

Find the acute angle between the lines  = (ˆi + 2ˆj + 4ˆk ) + t(2ˆi + 2ˆj + ˆk ) and the straight line passing through the points (5,1, 4) and (9, 2,12) .

Solution

We know that the line  = (ˆi + 2ˆj + 4ˆk ) + t(2ˆi + 2ˆj + ˆk ) is parallel to the vector 2ˆi + 2ˆj + ˆk.

Direction ratios of the straight line joining the two given points (5,1, 4) and (9, 2,12) are 4,1,8 and hence this line is parallel to the vector 4iˆ + ˆj + 8kˆ .

Therefore, the acute angle between the given two straight lines is


 

Example 6.30

Find the acute angle between the straight lines  and state whether they are parallel or perpendicular.

Solution

Comparing the given lines with the general Cartesian equations of straight lines,


we find (b1 , b2 , b3 ) = (2,1, −2) and (d1 , d2 , d3 ) = (4, −4, 2) . Therefore, the acute angle between the two straight lines is


Thus the two straight lines are perpendicular.

 

Example 6.31

Show that the straight line passing through the points A(6, 7, 5) and B(8,10, 6) is perpendicular to the straight line passing through the points C(10, 2, −5) and D(8, 3, −4) .

Solution

The straight line passing through the points A(6, 7, 5)  and B(8,10, 6) is parallel to the vector    −  = 2iˆ + 3 ˆj + kˆ and the straight line passing through the points C(10, 2, −5) and D(8, 3, −4) is parallel to the vector   = −2iˆ + ˆj + kˆ . Therefore, the angle between the two straight lines is the angle between the two vectors  and . Since


the two vectors are perpendicular, and hence the two straight lines are perpendicular.

Aliter

We find that direction ratios of the straight line joining the points A(6, 7, 5) and B(8,10, 6) are (b1 , b2 , b3 ) = (2, 3,1) and and direction ratios of the line joining the points C(10, 2, −5) and D(8, 3, −4) are (d1 , d2 , d3 ) = (−2,1,1) . Since b1d1 + b2d2 + b3d3 = (2)(−2) + (3)(1) + (1)(1) = 0 , the two straight lines are perpendicular.

 

Example 6.32

Show that the lines  and   are parallel

Solution

We observe that the straight line  is parallel to the vector 4iˆ - 6 ˆj +12kˆ and the straight line  is parallel to the vector -2iˆ + 3ˆj - 6kˆ.

Since 4iˆ - 6ˆj +12kˆ = -2(-2iˆ + 3ˆj - 6kˆ) , the two vectors are parallel, and hence the two straight lines are parallel.

 

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12th Mathematics : UNIT 6 : Applications of Vector Algebra : Application of Vectors to 3-Dimensional Geometry | Definition, Theorem, Proof, Solved Example Problems, Solution

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12th Mathematics : UNIT 6 : Applications of Vector Algebra


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