Advertisements
Advertisements
Question
O is the origin and A is (a, b, c). Find the direction cosines of the line OA and the equation of plane through A at right angle to OA.
Advertisements
Solution
Given, O(0, 0, 0) and A(a, b, c)
So, the direction ratios of OA = a – 0, b – 0, c – 0 = a, b, c
And, the direction cosines of line OA
`(a/(sqrt(a^2 + b^2 + c^2)), b/sqrt(a^2 + b^2 + c^2), c/sqrt(a^2 + b^2 + c^2))`
Now, the direction ratios of the normal to the plane are (a, b, c).
We know that, the equation of the plan passing through the point A(a, b, c) is
a(x – a) + b(y – b) + c(z – c) = 0
ax – a2 + by – b2 + cz – c2 = 0
ax + by + cz = a2 + b2 + c2
Thus, the required equation of the plane is ax + by + cz = a2 + b2 + c2
APPEARS IN
RELATED QUESTIONS
Name the octants in which the following points lie:
(7, 4, –3)
Name the octants in which the following points lie:
(–5, –3, –2)
Find the image of:
(5, 2, –7) in the xy-plane.
Find the image of:
(–5, 0, 3) in the xz-plane.
Find the image of:
(–4, 0, 0) in the xy-plane.
Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, –4).
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.
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.
Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
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 (x1, y1, z1) and (x2, y2, z2) 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 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.
Find the ratio in which the line segment joining the points (2, 4,5) and (3, −5, 4) is divided by the yz-plane.
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
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 from a point P(6,7, 8) on x - axis are
The x-coordinate of a point on the line joining the points Q(2, 2, 1) and R(5, 1, –2) is 4. Find its z-coordinate.
Find the image of the point having position vector `hati + 3hatj + 4hatk` in the plane `hatr * (2hati - hatj + hatk)` + 3 = 0.
A line makes equal angles with co-ordinate axis. Direction cosines of this line are ______.
If a line makes angles `pi/2, 3/4 pi` and `pi/4` with x, y, z axis, respectively, then its direction cosines are ______.
If a line makes angles α, β, γ with the positive directions of the coordinate axes, then the value of sin2α + sin2β + sin2γ is ______.
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.
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
Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).
Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 0
Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′, respectively, from the origin, prove that
`1/a^2 + 1/b^2 + 1/c^2 = 1/(a"'"^2) + 1/(b"'"^2) + 1/(c"'"^2)`
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 cartesian equation of the plane `vecr * (hati + hatj - hatk)` is ______.
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))`.
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.
