मराठी

Find the Image Of: (–4, 0, 0) in the Xy-plane.

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प्रश्न

Find the image  of: 

 (–4, 0, 0) in the xy-plane. 

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उत्तर

(-4,0,0)

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पाठ 28: Introduction to three dimensional coordinate geometry - Exercise 15.1 [पृष्ठ ६]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 28 Introduction to three dimensional coordinate geometry
Exercise 15.1 | Q 2.5 | पृष्ठ ६

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).


The x-axis and y-axis taken together determine a plane known as_______.


Name the octants in which the following points lie: (5, 2, 3)


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. 


Determine the points in zx-plane are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1). 


Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, –4).


Prove that the point A(1, 3, 0), B(–5, 5, 2), C(–9, –1, 2) and D(–3, –3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.


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.


Are the points A(3, 6, 9), B(10, 20, 30) and C(25, –41, 5), the vertices of a right-angled triangle?


Verify the following: 

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


Verify the following: 

(0, 7, 10), (–1, 6, 6) and (–4, 9, –6) are vertices of a right-angled 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 length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.


What is the locus of a point for which y = 0, z = 0?


The ratio in which the line joining (2, 4, 5) and (3, 5, –9) is divided by the yz-plane is


The ratio in which the line joining the points (a, b, c) and (–a, –c, –b) is divided by the xy-plane is


XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio


What is the locus of a point for which y = 0, z = 0?


The length of the perpendicular drawn from the point P (3, 4, 5) on y-axis 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 direction cosines of the line passing through the points P(2, 3, 5) and Q(–1, 2, 4).


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


A line makes equal angles with co-ordinate axis. Direction cosines of this line are ______.


Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).


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 area of the quadrilateral ABCD, where A(0, 4, 1), B(2,  3, –1), C(4, 5, 0) and D(2, 6, 2), is equal to ______.


The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is ______.


The cartesian equation of the plane `vecr * (hati + hatj - hatk)` is ______.


The unit vector normal to the plane x + 2y +3z – 6 = 0 is `1/sqrt(14)hati + 2/sqrt(14)hatj + 3/sqrt(14)hatk`.


The intercepts made by the plane 2x – 3y + 5z +4 = 0 on the co-ordinate axis are `-2, 4/3, - 4/5`.


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.


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