Advertisements
Advertisements
Question
Find the image of the point having position vector `hati + 3hatj + 4hatk` in the plane `hatr * (2hati - hatj + hatk)` + 3 = 0.
Advertisements
Solution
Let the given point be P`(hati + 3hatj + 4hatk)` and Q be the image of P in the plane `hatr * (2hati - hatj + hatk)` + 3 = 0 as shown in the figure.
Then PQ is the normal to the plane.
Since PQ passes through P and is normal to the given plane
So the equation of PQ is given by `vecr = (hati + 3hatj + 4hatk) + lambda(2hati - hatj + hatk)`
Since Q lies on the line PQ, the position vector of Q can be expressed as `(hati + 3hatj + 4hatk) + lambda(2hati - hatj + hatk)`
i.e., `(1 + 2lambda)hati + (3 - lambda)hatj + (4 + lambda)hatk`
Since R is the mid point of PQ, the position vector of R is
`([(1 + 2lambda)hati + (3 - lambda)hatj + (4 + lambda)hatk] + [hati + 3hatj + 4hatk])/2`
i.e., `(lambda + 1)hati + (3 - lambda/2)hatj + (4 + lambda/2)hatk`
Again, since R lies on the plane `vecr * (2hati - hatj + hatk)` + 3 = 0
We have `{(lambda + 1)hati + (3 - lambda/2)hatj + (4 + lambda/2)hatk} * (2hati - hatj + hatk) + 3` = 0
⇒ λ = –2
Hence, the position vector of Q is `(hati + 3hatj + 4hatk) - 2(2hati - hatj + hatk)`
i.e. `-3hati + 5hatj + 2hatk`.
APPEARS IN
RELATED QUESTIONS
Find the image of:
(–2, 3, 4) in the yz-plane.
Find the image of:
(5, 2, –7) in the xy-plane.
Find the image of:
(–5, 0, 3) in the xz-plane.
A cube of side 5 has one vertex at the point (1, 0, –1), and the three edges from this vertex are, respectively, parallel to the negative x and y axes and positive z-axis. Find the coordinates of the other vertices of the cube.
Find the distances of the point P(–4, 3, 5) from the coordinate axes.
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.
Determine the points in zx-plane are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).
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.
Find the coordinates of the point which is equidistant from the four points O(0, 0, 0), A(2, 0, 0), B(0, 3, 0) and C(0, 0, 8).
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:
(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Show that the points A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.
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 \[- \frac{a x_1 + b y_1 + c z_1 + d}{a x_2 + b y_2 + c z_2 + d}\]
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
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
If a line makes an angle of `pi/4` with each of y and z axis, then the angle which it makes with x-axis is ______.
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 angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 0
Find the length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 0.
Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
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
Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.
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 `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is ______.
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.
