Advertisements
Advertisements
Question
The coordinates of a point are (3, –2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.
Advertisements
Solution
The seven coordinates are as follows:
\[\left( - 3, 2, 5 \right)\]
\[\left( 3, 2, - 5 \right)\]
\[\left( - 3, - 2, 5 \right)\]
\[\left( 3, - 2, - 5 \right)\]
\[\left( - 3, 2, - 5 \right)\]
\[\left( - 3, - 2, - 5 \right)\]
\[\left( 3, 2, 5 \right)\]
APPEARS IN
RELATED QUESTIONS
The x-axis and y-axis taken together determine a plane known as_______.
Name the octants in which the following points lie:
(–5, 4, 3)
Name the octants in which the following points lie:
(4, –3, 5)
Name the octants in which the following points lie:
(–5, –4, 7)
Name the octants in which the following points lie:
(–5, –3, –2)
Find the image of:
(–4, 0, 0) in the xy-plane.
Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).
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.
Find the locus of P if PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (–1, 3, –7).
Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.
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 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.
Find the equation of the set of the points P such that its distances from the points A(3, 4, –5) and B(–2, 1, 4) are equal.
Find the ratio in which the sphere x2 + y2 + z2 = 504 divides the line joining the points (12, –4, 8) and (27, –9, 18).
Find the point on y-axis which is at a distance of \[\sqrt{10}\] units from the point (1, 2, 3).
Find the point on x-axis which is equidistant from the points A (3, 2, 2) and B (5, 5, 4).
The ratio in which the line joining (2, 4, 5) and (3, 5, –9) is divided by the yz-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 perpendicular distance of the point P(3, 3,4) from the x-axis is
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).
Find the image of the point having position vector `hati + 3hatj + 4hatk` in the plane `hatr * (2hati - hatj + hatk)` + 3 = 0.
If α, β, γ are the angles that a line makes with the positive direction of x, y, z axis, respectively, then the direction cosines of the line are ______.
Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.
Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).
Find the equations of the line passing through the point (3,0,1) and parallel to the planes x + 2y = 0 and 3y – z = 0.
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.
The sine of the angle between the straight line `(x - 2)/3 = (y - 3)/4 = (z - 4)/5` and the plane 2x – 2y + z = 5 is ______.
The direction cosines of the vector `(2hati + 2hatj - hatk)` are ______.
The vector equation of the line through the points (3, 4, –7) and (1, –1, 6) 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 angle between the line `vecr = (5hati - hatj - 4hatk) + lambda(2hati - hatj + hatk)` and the plane `vec.(3hati - 4hatj - hatk)` + 5 = 0 is `sin^-1(5/(2sqrt(91)))`.
The angle between the planes `vecr.(2hati - 3hatj + hatk)` = 1 and `vecr.(hati - hatj)` = 4 is `cos^-1((-5)/sqrt(58))`.
The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is `vecr = (5hati - 4hatj + 6hatk) + lambda(3hati + 7hatj - 2hatk)`.
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.
