मराठी

If l1, m1, n1 ; l2, m2, n2 ; l3, m3, n3 are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to l1 + l2 + l3, m1 + m2 + m3,

Advertisements
Advertisements

प्रश्न

If l1, m1, n1 ; l2, m2, n2 ; l3, m3, n3 are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to l1 + l2 + l3, m1 + m2 + m3, n1 + n2 + n3 makes equal angles with them.

बेरीज
Advertisements

उत्तर

Let direction vector of three mutually perpendicular lines be

`veca = l_1hati + m_1hatj + n_1hatk`

`vecb = l_2hati + m_2hatj + n_2hatk`

`vecc = l_3hati + m_3hatj + n_3hatk`

Let the direction vector associated with directions cosines l1 + l2 + l3, m1 + m2 + m3, n1 + n2 + n3 be

`vecp = (1_1 + l_2 + l_3)hati + (m_1 + m_2 + m_3)hati + (n_1 + n_2 + n_3)hatk`

As, lines associated with direction vectors a, b and c are mutually perpendicular

We have `veca . vecb`  = 0   ......[dot product of two perpendicular vector is 0]

⇒ l1l2 + m1m2 + n1n2 = 0   .....[1]

Similarly, `veca . vecc` = 0

⇒ l1l3 + m1m3 + n1n3 = 0   ......[2]

And `vecb . vecc` = 0

⇒ l2l3 + m2m3 + n2n3 = 0   ......[3]

Now, let x, y, z be the angles made by direction vectors a, b and c respectively with p

Therefore, `cosx = veca. vecp`

⇒ cos x = l1(l1 + l2 + l3) + m1(m1 + m2 + m3) + n1(n1+ n2 + n3)

⇒ cos x = l12 + l1l2 + l1l3 + m12 + m1m2 + m1m3 + n12 + n1n2 + n1n3

⇒ cos x = l12 + m12 + n12 + (l1l2 + m1m2 + n1n2) + (l1l3 + m1m3 + n1n3)

As we know l12 + m12 + n12 = 1 because sum of squares of direction cosines of a line is equal to 1

⇒ cos x = 1 + 0 = 1   ......[From, 1 and 2]

Similarly, cos y = 1 and cos z = 1

⇒ x = y = z = 0

Hence, angle made by vector p, with vectors a, b and c are equal!

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Introduction to Three Dimensional Geometry - Exercise [पृष्ठ २३७]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 12 Introduction to Three Dimensional Geometry
Exercise | Q 28 | पृष्ठ २३७

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Name the octants in which the following points lie:

(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (–4, 2, –5), (–4, 2, 5),

(–3, –1, 6), (2, –4, –7).


Coordinate planes divide the space into ______ octants.


Name the octants in which the following points lie: 

(–5, –4, 7) 


Name the octants in which the following points lie:

 (2, –5, –7) 


Find the image  of: 

 (–2, 3, 4) in the yz-plane.


Find the image  of: 

 (–5, 4, –3) in the xz-plane. 


Find the image  of: 

 (–5, 0, 3) in the xz-plane. 


Planes are drawn through the points (5, 0, 2) and (3, –2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelepiped so formed. 


Find the distances of the point P(–4, 3, 5) from the coordinate axes. 


Find the points on z-axis which are at a distance \[\sqrt{21}\]from the point (1, 2, 3). 


Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.


Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of a square.


If A(–2, 2, 3) and B(13, –3, 13) are two points.
Find the locus of a point P which moves in such a way the 3PA = 2PB.


Find the locus of P if PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (–1, 3, –7).


Verify the following: 

 (0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle.


Verify the following:

 (5, –1, 1), (7, –4,7), (1, –6,10) and (–1, – 3,4) are the vertices of a rhombus.


Find the locus of the point, the sum of whose distances from the points A(4, 0, 0) and B(–4, 0, 0) is equal to 10.


Show that the plane ax + by cz + d = 0 divides the line joining the points (x1y1z1) and (x2y2z2) in the ratio \[- \frac{a x_1 + b y_1 + c z_1 + d}{a x_2 + b y_2 + c z_2 + d}\]


Write the distance of the point P(3, 4, 5) from z-axis.


Write the coordinates of the foot of the perpendicular from the point (1, 2, 3) on y-axis.


Write the length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.


Let (3, 4, –1) and (–1, 2, 3) be the end points of a diameter of a sphere. Then, the radius of the sphere is equal to 


The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz- plane are


The coordinates of the foot of the perpendicular from a point P(6,7, 8) on x - axis are 


The perpendicular distance of the point P (6, 7, 8) from xy - plane is


The perpendicular distance of the point P(3, 3,4) from the x-axis is 


If the direction ratios of a line are 1, 1, 2, find the direction cosines of the line.


Find the co-ordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1).


The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by ______.


If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane


The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is ax + by `+- (sqrt(a^2 + b^2) tan alpha)z ` = 0


The line `vecr = 2hati - 3hatj - hatk + lambda(hati - hatj + 2hatk)` lies in the plane `vecr.(3hati + hatj - hatk) + 2` = 0.


If the foot of perpendicular drawn from the origin to a plane is (5, – 3, – 2), then the equation of plane is `vecr.(5hati - 3hatj - 2hatk)` = 38.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×