1. Scalar Product of Vectors

Notes:
  1. We can perform addition, subtraction and multiplication of vector quantities. However, division of them is not possible.
  2. During multiplication, we may get a scalar result or a vector result depending upon the type of multiplication we choose.
  3. There are two types of multiplication of vectors: Dot Product and Cross Product.
  4. In class 10, we have discussed only Dot Product and this always gives us a scalar quantity as a result.
  5. $\vec{a}.\vec{b} = |\vec{a}|.|\vec{b}| cos \theta$
Important:
  1. When $\theta$ = 90°, $\vec{a}.\vec{b}=0$

1. Find the scalar product of given pair of vectors:

a) $\vec{a} = 4\vec{i} + 6\vec{j}$ and $\vec{b} = 6\vec{i} + 7 \vec{j}$
Solution:
Here,
$\vec{a} = 4\vec{i} + 6\vec{j} = (4,6)$

$\vec{b} = 6\vec{i} + 7 \vec{j} = (6,7)$

$\vec{a}.\vec{b} = (4*6+ 6*7)$

$= (24+42)$
$= 66$


c) $\vec{a} = 5\vec{i} + 3\vec{j}$ and $\vec{b} = 2\vec{i} -4 \vec{j}$
Solution:

Here,
$\vec{a} = 5\vec{i} + 3\vec{j} = (5,3)$

$\vec{b} =2\vec{i} -4 \vec{j} = (2,-4)$

$\vec{a}.\vec{b} = 5*2+ 3*(-4)$

$= 10-12$

$= -2$



e) $\vec{a} = 3\vec{i} + 8\vec{j}$ and $\vec{b} = 5\vec{i} + 12 \vec{j}$
Solution:

Here,
$\vec{a} = 3\vec{i} + 8\vec{j} = (3,8)$

$\vec{b} = 5\vec{i} + 12 \vec{j} = (5,12)$

$\vec{a}.\vec{b} = (3*5+ 8*12)$

$= (15+96)$

$= 111$



g) $\vec{a} = \left ( 6 \\ -4 \right )$ and $\vec{b} = \left ( -3 \\ -5 \right)$

Solution:

$\vec{a}.\vec{b} = \left ( 6 \\-4 \right) . \left ( -3 \\ -5 \right)$

$= 6(-3) + (-4)(-5)$

$= -18 + 20$

$= 2$





About Readmore Optional Mathematics Book 10

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Editor: I. R. Simkhada

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Class 10 - Scalar Product of Vectors Solved Exercises | Readmore Optional Mathematics is a collection of the solutions related to proofs of scalar product of vectors from the vector chapter for Nepal's Secondary Education Examination (SEE) appearing students.

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