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
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_______.
Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) and C (–1, 1, 2). Find the coordinates of the fourth vertex.
If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of a, b and c.
Name the octants in which the following points lie:
(4, –3, 5)
Name the octants in which the following points lie:
(2, –5, –7)
Find the image of:
(–4, 0, 0) in the xy-plane.
Planes are drawn parallel to the coordinate planes through the points (3, 0, –1) and (–2, 5, 4). Find the lengths of the edges of the parallelepiped so formed.
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.
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 point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).
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.
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.
Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.
Verify the following:
(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
Verify the following:
(5, –1, 1), (7, –4,7), (1, –6,10) and (–1, – 3,4) are the vertices of a rhombus.
Find the ratio in which the sphere x2 + y2 + z2 = 504 divides the line joining the points (12, –4, 8) and (27, –9, 18).
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 length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.
Find the point on y-axis which is at a distance of \[\sqrt{10}\] units from the point (1, 2, 3).
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
Find the coordinates of the point where the line through (3, – 4, – 5) and (2, –3, 1) crosses the plane passing through three points (2, 2, 1), (3, 0, 1) and (4, –1, 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 the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).
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.
Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
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 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 plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis. The value of α is equal to ______.
The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is ______.
The intercepts made by the plane 2x – 3y + 5z +4 = 0 on the co-ordinate axis are `-2, 4/3, - 4/5`.
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)))`.
