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Multiplication of a Vector by a Scalar

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Multiplication of a Vector by a Scalar - Lesson 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 → = (a1 – b1) i ^ + (a2 – b2) j ^ + (a3 – b3) 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 →   = (x2 – x1) i ^ + (y2 – y1) j ^ + (z2 – z1) k ^

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

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