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The scalar equals :
if any two vector are equal out of
= 0
For nonzero vectors holds if and only if
where θ is angle between
where θ is angle between
The volume of the parallelopiped whose sides are given by
Vol. of parallelopiped =
The points with position vectors 60i + 3j, 40 i – 8 j, ai – 52 j are collinear if
Three pts A, B, C are collinear if
Let be three non  coplanar vectors and are vectors defined by the relations then the value of the expression is equal to
Given that are non coplanar
Let a, b, c be distinct nonnegative numbers. If the vectors lie in a plane, then c is
a, b, c are distinct non negative numbers and the vectors are coplanar.
⇒ c^{2} – ac – ab + ac = 0
⇒ c^{2} = ab Þ a, c, b are in G.P.
∴ c is the G.M. of a and b.
Let be the position vectors of P and Qr espectively, with respect to O and The points R and S divide PQ internally and externally in the ratio 2 : 3 respectively. If OR and OS are perpendicular then
We have
[Internal divison]
[external divison]
Let α, β, γ be distinct real numbers. The points with position vectors
Let the given position vectors be of point A, B and C respectively, then
⇒ ΔABC is an equilateral Δ.
Let is a unit vector such that equals
where x^{2} + y^{2} + z^{2} = 1 ....(1)
⇒ x – y = 0 ⇒ x = y ...(2)
⇒ x + y + z = 0
⇒ 2x + z = 0 (using (2))
⇒ z = – 2x ...(3) From (1), (2) and (3)
If are non coplanar unit vectors such that then the angle between is
are noncoplanar]
Let be vectors such that If and
If are three non coplanar vectors, then equals
Let a = 2i + j – 2k and b = i + j. If c is a vector such that a. c =  c ,  c  a  = 2√2 and the angle between (a × b) and c is 30°, then  (a × b) × c =
Substituting values of in (1), we get
Let a =2i + j + k, b = i +2j –k and a unit vector c be coplanar. If c is perpendicular to a, then c =
As c is coplanar with a and b, we take,
c = αa + βb ...(1)
where α,β are scalars.
As c is perpendicular to a , c.a = 0
∴ From (1) we get, 0 = α a.a + β b.a
If the vectors form the sides BC, CA and ABrespectively of a triangle ABC, then
(by triangle law)
Let the vectors be such that Let P_{1} and P_{2} be planes determined by the pairs of vectors respectively. Thenthe angle between P_{1} and P_{2} is
Given that are vectors such that
P_{1} is the plane determined by vectors
∴ Normal vectors will be given by
Similarly, P_{2} is the plane determined by vectors
∴ Normal vectors to P_{2} will be given by
Substituting the values of
and hence the planes will also be parallel to each other.
Thus angle between the planes = 0
If are unit coplanar vectors, then the scalar triple product
are unit coplanar vectors, and a re a lso coplan a r vector s. bein g lin ea r combination of
and depends on
= 1(1+ x  y  x + x^{2}) 1(x^{2}y) = 1
∴ Depends neither on x nor on y.
If are unit vectors, then does NOT exceed
are two unit vectors such that and are perpendicular to each other then the angle between
Given that are two unit vectors
Let is a unit vector,, then the maximum value of the scalar triple product
Given that and u is a unit vector
which is max. when cos θ =1
The value of k such that lies in the plane 2x – 4y + z = 7, is
As the line lies in th e plan e 2x  4 y +z= 7, the point (4, 2, k) through which line passes must also lie on the given plane and hence 2 × 4 – 4 × 2 + k = 7 ⇒ k = 7
The value of ‘a’ so that the volume of parallelopiped formed by becomes minimum is
Vol. of parallelopiped formed by
= 1(1 0)  a(0  a^{2}) +1(0a) = 1 + a^{3}a
If the lines
intersect, then the value of k is
⇒ x = 3 + m,y = k +2μ and z = μ
Since above lines intersect
...(1)
3λ  1 = 2μ + k ...(2)
μ = 4λ+1 ..(3)
Solving (1) and (3) and putting the value of l and m in (2) we get, k = 9/2
The unit vector which is orthogonal to the vector an d is coplanar with the vectors and
A variable plane at a distance of the one unit from the origin cuts the coordinates axes at A, B and C. If the centroid D (x, y, z) of triangle ABC satisfies the relation , then the value k is
Let the eq^{n} of variable plane be which meets the axes at A (a, 0, 0), B (0, b, 0) and C (0. 0, c).
∴ Centroid of
and it satisfies the relation
...(1)
Also given that the distance of plane from (0, 0, 0) is 1 unit.
From (1) and (2), we get
If are three nonzero, noncoplanar vectors and
then the set of orthogonal vectors is
We observe that
is a set of orthogonal vectors.
A plane which is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4, passes through (1, –2, 1). The distance of the plane from the point (1, 2, 2) is
The equation of plane through the point (1, –2, 1) and perpendicular to the planes 2 x  2 y +z=0 and x  y + 2z=4 is given by
It’s distance from the point (1, 2, 2) is
L et A vector in the plane of whose projection on
A vector in the plane of is
Projection of
The number of distinct real values of λ, for which the vectors are coplanar, is
We know that three vector are coplanar if their scalar triple product is zero.
∴ Two real solutions.
Let be unit vectors such that Which one of the following is correct ?
Since, ar e unit vectors ,
therefore form an equilateral triangle.
Also since are non parallel (these form an equilateral Δ).
The edges of a parallelopiped are of unit length and are parallel to noncoplan ar unit vectors such that Then, the volume of the parallelopiped is
We know that th e volume of a par allelopipe with coterminus edges as the vectors is given by
Let two noncollinear unit vectors form an acute angle. A point P moves so that at any time t the position vector (where O is the origin) is given by When P is farthest from origin O, let M be the length of and be the unit vector along
Then,
Maximum occurs at
Let P (3, 2, 6) be a point in space and Q be a point on the line
Then the value of m for which the vector is parallel to the plane x – 4y + 3z = 1 is
Given that P (3, 2, 6) is a point in space and Q is a point on line
are unit vectors such that
are coplanar
So, angle between should also be π / 3.
Hence are non parallel.
A line with positive direction cosines passes through the point P(2, –1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals
The line has +ve and equal direction cosines, these are or direction ratios are 1, 1, 1. Also thelines passes through P (2, – 1, 2).
∴ Equation of line is
be a point on this line where it meets the plane 2 x + y + z = 9
Then Q must satisfy the eq^{n} of plane
∴ Q has coordintes (3, 0, 3)
Let P, Q, R and S be the points on the plane with position vectors respectively. The quadrilateral PQRS must be a
⇒ PQRS is a parallelogram but neither a rhombus nor a rectangle.
Equation of the plane containing the straight line and perpendicular to the plane containing the
straight lines
Plane containing two lines and
Now equation of plane containing the line and perpendicular to the plane 8x – y – 10 z = 0 is
⇒ –26x +52y –26z = 0 or x –2y+ z = 0
If the distance of the point P (1, –2, 1) from the plane x + 2y –2z = α, where α > 0, is 5, then the foot of the perpendicular from P to the plane is
As perpendicular distance of x + 2 y –2z = a from the point (1, – 2, 1) is 5
∴ Plane becomes x + 2 y – 2z – 10 = 0
Two adjacent sides of a parallelogram ABCD are given by
The side AD is rotated by an acute angle a in the plane of the parallelogram so that AD becomes AD¢. If AD¢ makes a right angle with the side AB, then the cosine of the angle a is given by
Let be three vector s. A vector the plane of whose projection on , is given by
The point P is the intersection of the straight line joining the points Q(2, 3, 5) and R(1, –1, 4) with the plane 5x – 4y – z = 1. If S is the foot of the perpendicular drawn from the point T(2, 1, 4) to QR, then the length of the line segment PS is
Equation of st. line join ing Q (2, 3, 5) an d R (1, –1, 4) is
Now let point S on QR be
∵ S is the foot of perpendicular drawn from T (2, 1, 4) to QR, where dr’s of ST are µ, 4µ – 2, µ –1 and dr’s of QR are –1, – 4, –1
∴ Distance between P and S
The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x – y + z = 3 and at a distance from the point (3, 1 ,–1) is
The plane passing thr ough the intersection line of given planes is
(x + 2y + 3z – 2) + λ( x y +z  3)=0 or (1 +λ)x + (2 λ)y + (3 +λ)z + (2  3λ) = 0
Its distance from the point (3, 1, –1) is
∴ Required equation of plane is
or 5x – 11y + z = 17
If are vectors such that and then a possible value of
Given that
But neither is a null vector
Let P be the image of the point (3,1,7) with r espect to the plane x – y + z = 3. Then the equation of the plane passing through P and containing the straight line
P, the image of point (3, 1, 7) in the plane x – y + z = 3 is given by
Now equation of plane through (–1, 5, 3) and containing the
line
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