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 **R**^{3}. 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 **R**^{3}
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.

A straight line can be uniquely fixed if

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

Â·
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
â€“ parametric form**.

**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
*âˆˆ **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*.

Since * *is parallel to * *, we have * *Ã— * *=

That is, (* *âˆ’ ) Ã— * *= 0 .

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

Suppose *P *is (*x*, *y*, *z*) , *A *is
(*x*_{1} , *y*_{1} , *z*_{1} ) and * *=
*b*_{1} Ë†*i*
+ *b*_{2} Ë†*j *+ *b*_{3} Ë†*k *. Then, substituting * *= *x* Ë†*i *+ *y* Ë†*j *+ *z* Ë†*k *, * *= *x*_{1}Ë†*i *+ *y*_{1}Ë†* j *+ *z*_{1} Ë†*k* in (1) and comparing
the coefficients of Ë†*i *,
Ë†*j*, Ë†*k *, we get

*x *âˆ’ *x*_{1} = *tb*_{1} , *y *âˆ’ *y*_{1}
= *tb*_{2} , *z *âˆ’ *z*_{1} = *tb*_{3 } â€¦â€¦â€¦â€¦.(4)

Conventionally (4) can be written as

which are called the **Cartesian equations or symmetric equations **of a straight line
passing through the point (*x*_{1}, *y*_{1} , *z*_{1})
and parallel to a vector with direction ratios *b*_{1}, *b*_{2},
*b*_{3}.

(i) Every point on the line (5) is of the form
(*x*_{1} + *tb*_{1} , *y*_{1} + *tb*_{2}
, *z*_{1} + *tb*_{3}) , where *t *âˆˆ **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 *b*_{1}, *b*_{2}
, *b*_{3} are zero, it does not mean that we are dividing by zero.
But it means that the corresponding numerator is zero. For instance, If *b*_{1}
â‰ 0, *b*_{2} â‰ 0 and *b*_{3} = 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.

**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*_{2} âˆ’ *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 - x _{1})
/ b_{1} = y - y_{1} / b_{1} = (z - z_{1}) / b_{1}.*

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 = b_{1}*i*Ë† + b_{2}Ë†*j* + b_{3}*k*Ë†
, then
direction ratios of the straight
line are b_{1} , b_{2}
, b_{3}. 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 (x_{1} , y_{1} , z_{1}
) = (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.

The acute angle between two given straight lines

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

**Remark**

(i) The two given lines with direction ratios b_{1} , b_{2}
, b_{3 }and d_{1} , d_{2} , d_{3 }are parallel
if, and only if .

(ii) The two given lines
with direction ratios b_{1}, b_{2}, b_{3} and d_{1},
d_{2}, d_{3} are perpendicular if and only if *b*_{1}*d*_{1} + *b*_{2}*d*_{2} + *b*_{3}*d*_{3}
= 0 .

(iii) If the direction
cosines of two given straight lines are l_{1} , m_{1} , n_{1}
and l_{2} , m_{2} , n_{2}, then the angle between the
two given straight lines is cos Î¸ =| l_{1}l_{2} + m_{1}m_{2}
+ n_{1}n_{2} | .

Find the acute angle between the lines * *= (Ë†*i *+ 2Ë†*j *+ 4Ë†*k *) + *t*(2Ë†*i *+ 2Ë†*j *+ Ë†*k *) and the straight line
passing through the points (5,1, 4) and (9, 2,12) .

We know that the line * *= (Ë†*i *+ 2Ë†*j *+ 4Ë†*k *) + *t*(2Ë†*i *+ 2Ë†*j *+ Ë†*k *) is parallel to the
vector 2Ë†*i *+ 2Ë†*j *+ Ë†*k*.

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
4*i*Ë† + Ë†*j *+ 8*k*Ë† .

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

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

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

we find (*b*_{1} , *b*_{2} , *b*_{3}
) = (2,1, âˆ’2) and (*d*_{1} , *d*_{2} , *d*_{3}
) = (4, âˆ’4, 2) . Therefore, the acute angle between the two straight lines is

Thus the two straight lines are perpendicular.

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, âˆ’5) and *D*(8, 3, âˆ’4) .

The straight line passing through the points* A*(6, 7, 5)
and *B*(8,10, 6) is parallel to the vector * *= * *= * *âˆ’ * *= 2*i*Ë† + 3 Ë†*j *+ *k*Ë† and the straight
line passing through the points *C*(10, 2, âˆ’5) and *D*(8, 3, âˆ’4) is
parallel to the vector * *= * *= âˆ’2*i*Ë† + Ë†*j
*+ *k*Ë† . 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 (*b*_{1} , *b*_{2}
, *b*_{3} ) = (2, 3,1) and and direction ratios of the line joining the
points *C*(10, 2, âˆ’5) and *D*(8, 3, âˆ’4) are (*d*_{1} , *d*_{2}
, *d*_{3} ) = (âˆ’2,1,1) . Since *b*_{1}*d*_{1}
+ *b*_{2}*d*_{2} + *b*_{3}*d*_{3}
= (2)(âˆ’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 4*i*Ë† - 6 Ë†*j* +12*k*Ë†
and the straight line is parallel to the vector -2*i*Ë† + 3Ë†*j* - 6*k*Ë†.

Since 4*i*Ë† - 6Ë†*j* +12*k*Ë†
= -2(-2*i*Ë† + 3Ë†*j* - 6*k*Ë†) , the two
vectors are parallel, and hence the two straight lines are parallel.

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

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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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