Notes On Multiplication of a Vector by a Scalar - CBSE Class 12 Maths

If we multiply any vector by a scalar quantity we get another vector.

λ x  r   = λ r

|λ r | = |λ| x |  r |

r and λ r are collinear.

If λ = -1 then λ r = - r

r + (- r ) = 0

If λ =  1 r , r _ 0 _

λ r = 1 r r

λ r | = 1 r r   = 1

⇒ λ r = r ^ = 1 r r

For any two scalars, l and m, and any two vectors, a and  b :

l a + m  a = (l + m) a

l(ma ) = (lm)a

l(a + b ) = l a + l b

Let P be a point in space with coordinates x, y, z.

Let us draw a perpendicular, PA, from point P on to the XOY plane.

Let Q and S be the feet of the perpendiculars drawn from point A to the X and Y axes, respectively.

Now, Q, S and R are the points on the X, Y and Z axes, respectively, such that OQ is equal to X, OS is equal to Y and OR is equal to Z.

Now, vectors I, J and K are unit vectors along the X, Y and Z axes, respectively.

OQ = x i ^

OS = y j ^

OR = z k ^

From the figure:

QA = OS = y j ^

In ∆ OQA:

OA = OQ + QA [Triangle law of vector addition]

OA = x i ^ + y j ^

From the figure:

AP = OR = z k ^

In ∆ OAP:

OP = OA + AP [Triangle law of vector addition]

OP = x i ^ + y j ^ + z k ^

Component form of vector OP :

OP = x i ^ + y j ^ + z k ^

In right angled ∆ OQA:

OA 2 = OQ 2 + QA 2

⇒ ⃓ OA 2 = x2 + y2

In right angled ∆ OAP:

OP 2 OA 2 AP 2

⇒ ⃓ OP 2 = x2 + y2 + z2

⇒  OP = x 2 + y 2 + z 2

Direction cosines of OP :

l = cos α

m = cos β

n = cos γ

Unit vector  a ^ in the direction of OP :

a ^ = l i ^ + m j ^ + n k ^ = cos α i ^ + cos β  j ^ + cos ɣ  k ^

a = a1 i ^ + a2 j ^ + a3 k ^

λ a = λ a1 i ^ + λ a2 j ^ + λ a3 k ^

b = b1 i ^ + b2 j ^ + b3 k ^

a + b = (a1 + b1) i ^ + (a2 + b2) j ^ + (a3 + b3) k ^

a - b = (a1b1) i ^ + (a2 – b2) j ^ + (a3b3) k ^

a = b a1 = b1, a2 = b2 and a3 = b3

a and b are collinear if only if there exists a scalar, λ, such that:

a1 i ^ + a2 j ^ + a3 k ^ = λ ( b1 i ^ + b2 j ^ + b3 k ^ )

a1 i ^ + a2 j ^ + a3 k ^ =  λb1 i ^ + λb2 j ^ + λb3 k ^ ) ⇒a1 = λb1, a2 = λb2 and a3 = λb3

a 1 b 1 = a 2 b 2 = a 3 b 3 = λ

Component form ofVector Joining Two Points

Given A1 (x1, y1, z1) and A2 (x2, y2, z2).

In ∆OP1P2:

OP 2 = OP 1 + P 1 P 2   [By triangle law of vector addition]

OP 1 = OP 2 P 1 P 2   ....(1)

OP 1 = x1 i ^ + y1 j ^ + z1 k ^

OP 2 = x2 i ^ + y2 j ^ + z2 k ^

P 1 P 2  = ( x2 i ^ + y2 j ^ + z2 k ^ ) – (x1 i ^ + y1 j ^ + z1 k ^ )

P 1 P 2   = (x2x1) i ^ + (y2 – y1) j ^ + (z2z1) k ^

P 1 P 2   | = x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2

Summary

If we multiply any vector by a scalar quantity we get another vector.

λ x  r   = λ r

|λ r | = |λ| x |  r |

r and λ r are collinear.

If λ = -1 then λ r = - r

r + (- r ) = 0

If λ =  1 r , r _ 0 _

λ r = 1 r r

λ r | = 1 r r   = 1

⇒ λ r = r ^ = 1 r r

For any two scalars, l and m, and any two vectors, a and  b :

l a + m  a = (l + m) a

l(ma ) = (lm)a

l(a + b ) = l a + l b

Let P be a point in space with coordinates x, y, z.

Let us draw a perpendicular, PA, from point P on to the XOY plane.

Let Q and S be the feet of the perpendiculars drawn from point A to the X and Y axes, respectively.

Now, Q, S and R are the points on the X, Y and Z axes, respectively, such that OQ is equal to X, OS is equal to Y and OR is equal to Z.

Now, vectors I, J and K are unit vectors along the X, Y and Z axes, respectively.

OQ = x i ^

OS = y j ^

OR = z k ^

From the figure:

QA = OS = y j ^

In ∆ OQA:

OA = OQ + QA [Triangle law of vector addition]

OA = x i ^ + y j ^

From the figure:

AP = OR = z k ^

In ∆ OAP:

OP = OA + AP [Triangle law of vector addition]

OP = x i ^ + y j ^ + z k ^

Component form of vector OP :

OP = x i ^ + y j ^ + z k ^

In right angled ∆ OQA:

OA 2 = OQ 2 + QA 2

⇒ ⃓ OA 2 = x2 + y2

In right angled ∆ OAP:

OP 2 OA 2 AP 2

⇒ ⃓ OP 2 = x2 + y2 + z2

⇒  OP = x 2 + y 2 + z 2

Direction cosines of OP :

l = cos α

m = cos β

n = cos γ

Unit vector  a ^ in the direction of OP :

a ^ = l i ^ + m j ^ + n k ^ = cos α i ^ + cos β  j ^ + cos ɣ  k ^

a = a1 i ^ + a2 j ^ + a3 k ^

λ a = λ a1 i ^ + λ a2 j ^ + λ a3 k ^

b = b1 i ^ + b2 j ^ + b3 k ^

a + b = (a1 + b1) i ^ + (a2 + b2) j ^ + (a3 + b3) k ^

a - b = (a1b1) i ^ + (a2 – b2) j ^ + (a3b3) k ^

a = b a1 = b1, a2 = b2 and a3 = b3

a and b are collinear if only if there exists a scalar, λ, such that:

a1 i ^ + a2 j ^ + a3 k ^ = λ ( b1 i ^ + b2 j ^ + b3 k ^ )

a1 i ^ + a2 j ^ + a3 k ^ =  λb1 i ^ + λb2 j ^ + λb3 k ^ ) ⇒a1 = λb1, a2 = λb2 and a3 = λb3

a 1 b 1 = a 2 b 2 = a 3 b 3 = λ

Component form ofVector Joining Two Points

Given A1 (x1, y1, z1) and A2 (x2, y2, z2).

In ∆OP1P2:

OP 2 = OP 1 + P 1 P 2   [By triangle law of vector addition]

OP 1 = OP 2 P 1 P 2   ....(1)

OP 1 = x1 i ^ + y1 j ^ + z1 k ^

OP 2 = x2 i ^ + y2 j ^ + z2 k ^

P 1 P 2  = ( x2 i ^ + y2 j ^ + z2 k ^ ) – (x1 i ^ + y1 j ^ + z1 k ^ )

P 1 P 2   = (x2x1) i ^ + (y2 – y1) j ^ + (z2z1) k ^

P 1 P 2   | = x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2

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