Show that the vector i ^ + j ^ + k ^ is equally inclined to the coordinate axes.

Let θ 1 be the angle between a → and x - axis. | a → | = ( 1 ) 2 + ( 1 ) 2 + ( 1 ) 2 = 3 b → = i ^ .................. (Because i ^ is the unit vector along x -axis) | b → | = ( 1 ) 2 = 1 = 1 a → · b → = 1 + 0 + 0 = 1 cos θ 1 = a → · b → | a → | | b → | = 1 ( 3 ) ( 1 ) = 1 3 ⇒ θ 1 = cos - 1 1 3 . . . . . . . . . ( 1 ) Let θ 2 be the angle between a → and y -axis. | a → | = ( 1 ) 2 + ( 1 ) 2 + ( 1 ) 2 = 3 b → = j ^ .................. (Because j ^ is the unit vector along y -axis) | b → | = ( 1 ) 2 = 1 = 1 a → · b → = 0 + 1 + 0 = 1 cos θ 2 = a → · b → | a → | | b → | = 1 ( 3 ) ( 1 ) = 1 3 ⇒ θ 2 = cos - 1 1 3 . . . . . . . . . . . . . . ( 2 ) Let θ 3 be the angle between a → and z -axis. | a → | = ( 1 ) 2 + ( 1 ) 2 + ( 1 ) 2 = 3 b → = k ^ ........................ (Because k ^ is the unit vector along z -axis) | b → | = ( 1 ) 2 = 1 = 1 a → · b → = 0 + 0 + 1 = 1 cos θ = a → · b → | a → | | b → | = 1 ( 3 ) ( 1 ) = 1 3 ⇒ θ = cos - 1 1 3 . . . . . . . . . . . ( 3 ) From ( 1 ) , ( 2 ) and ( 3 ) , the given vector is equally inclined to the coordinate axes.

Show that the vectors a → = 1 7 ( 2 i ^ + 3 j ^ + 6 k ^ ) , b → = 1 7 ( 3 i ^ - 6 j ^ + 2 k ^ ) , c → = 1 7 ( 6 i ^ + 2 j ^ - 3 k ^ ) are mutually perpendicular unit vectors.

We have | a → | = 1 7 2 2 + 3 2 + 6 2 = 1 7 49 = 7 7 = 1 | b → | = 1 7 3 2 + ( - 6 ) 2 + 2 2 = 1 7 49 = 7 7 = 1 | c → | = 1 7 6 2 + 2 2 + ( - 3 ) 2 = 1 7 49 = 7 7 = 1 And a → · b → = 1 7 ( 2 i ^ + 3 j ^ + 6 k ^ ) · 1 7 ( 3 i ^ - 6 j ^ + 2 k ^ ) = 1 49 ( 6 - 18 + 12 ) = 0 b → · c → = 1 7 ( 3 i ^ - 6 j ^ + 2 k ^ ) · 1 7 ( 6 i ^ + 2 j ^ - 3 k ^ ) = 1 49 ( 18 - 12 - 6 ) = 0 c → · a → = 1 7 ( 6 i ^ + 2 j ^ - 3 k ^ ) · 1 7 ( 2 i ^ + 3 j ^ + 6 k ^ ) = 1 49 ( 12 + 6 - 18 ) = 0 So, | a → | = | b → | = | c → | = 1 and a → · b → = b → · c → = c → · a → = 0 So, the given vectors are mutually perpendicular unit vectors.

For any two vectors a → and b → , show that: ( a → + b → ) · ( a → - b → ) = 0 ⇔ | a → | = | b → | .

We have ( a → + b → ) · ( a → - b → ) = 0 ⇒ a → · a → - a → · b → + b → · a → - b → · b → = 0 ⇒ a → · a → - a → · b → + a → · b → - b → · b → = 0 (Using commutative property of vectors) ⇒ | a → | 2 - | b → | 2 = 0 ⇒ | a → | 2 = | b → | 2 ⇒ | a → | = | b → |

If p → = 5 i ^ + λ j ^ - 3 k ^ and q → = i ^ + 3 j ^ - 5 k ^ then find the value of λ , so that p → + q → and p → - q → are perpendicular vectors.

Given that p → = 5 i ^ + λ j ^ - 3 k ^ and q → = i ^ + 3 j ^ - 5 k ^ p → + q → = ( 5 i ^ + λ j ^ - 3 k ^ ) + ( i ^ + 3 j ^ - 5 k ^ ) = 6 i ^ + ( λ + 3 ) j ^ - 8 k ^ p → - q → = ( 5 i ^ + λ j ^ - 3 k ^ ) - ( i ^ + 3 j ^ - 5 k ^ ) = 4 i ^ + ( λ - 3 ) j ^ + 2 k ^ Given that p → + q → is orthogonal to p → - q → . ⇒ ( p → + q → ) · ( p → - q → ) = 0 ⇒ [ 6 i + ( λ + 3 ) j ^ - 8 k ^ ] · [ 4 i ^ + ( λ - 3 ) j ^ + 2 k ^ ] = 0 ⇒ 24 + λ 2 - 9 - 16 = 0 ⇒ λ 2 = 1 ∴ λ = ± 1

If α → = 3 i ^ + 4 j ^ + 5 k ^ and β → = 2 i ^ + j ^ - 4 k ^ , then express β → in the form of β → = β 1 → + β 2 → , where β 1 → is parallel to α → and β 2 → is perpendicular to α → .

Given that α → = 3 i ^ + 4 j ^ + 5 k ^ and β → = 2 i ^ + j ^ - 4 k ^ and β → = β 1 → + β 2 → ⇒ β 2 → = β → - β 1 → . . . . . . . . . . . ( 1 ) since β → 1 is parallel to α → , β 1 → = t α → ⇒ β 1 → = t ( 3 i ^ + 4 j ^ + 5 k ^ ) = 3 t i ^ + 4 t j ^ + 5 t k ^ . . . . . . . . . ( 2 ) Substituting the values of β 1 → and α → in ( 1 ) , we get β 2 → = 2 i ^ + j ^ - 4 k ^ - ( 3 t i ^ + 4 t j ^ + 5 t k ^ ) = ( 2 - 3 t ) i ^ + ( 1 - 4 t ) j ^ + ( - 4 - 5 t ) k ^ . . . . . . . . . . . . ( 3 ) since β 2 → is perpendicular to α → , β 2 → · α → = 0 ⇒ [ ( 2 - 3 t ) i ^ + ( 1 - 4 t ) j ^ + ( - 4 - 5 t ) k ^ ] · ( 3 i ^ + 4 j ^ + 5 k ^ ) = 0 ⇒ 3 ( 2 - 3 t ) + 4 ( 1 - 4 t ) + 5 ( - 4 - 5 t ) = 0 ⇒ 6 - 9 t + 4 - 16 t - 20 - 25 t = 0 ⇒ - 50 t = 10 ⇒ t = - 1 5 From ( 2 ) and ( 3 ) , we get β 1 → = - 1 5 ( 3 i ^ + 4 j ^ + 5 k ^ ) β 2 → = 13 5 i ^ + 9 5 j ^ - 3 k ^ = 1 5 ( 13 i ^ + 9 j ^ - 15 k ^ )

E M B I B E

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>EXERCISE>Q 9

MEDIUM

12th CBSE

IMPORTANT

Earn 100

If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is $\sqrt{3}$ .

Important Questions on Scalar or Dot Product

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>EXERCISE>Q 10

If

\vec{a}, \vec{b}, \vec{c}

are three mutually perpendicular unit vectors, then prove that

| \vec{a} + \vec{b} + \vec{c} | = \sqrt{3}

.

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>EXERCISE>Q 11

If

| \vec{a} + \vec{b} | = 60, | \vec{a} - \vec{b} | = 40

and

| \vec{b} | = 46

, find

| \vec{a} |

.

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>EXERCISE>Q 12

Show that the vector

\hat{i} + \hat{j} + \hat{k}

is equally inclined to the coordinate axes.

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>EXERCISE>Q 13

Show that the vectors $\vec{a} = \frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}), \vec{b} = \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}), \vec{c} = \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k})$ are mutually perpendicular unit vectors.

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>EXERCISE>Q 14

For any two vectors

\vec{a}

and

\vec{b}

, show that:

(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0 \Leftrightarrow | \vec{a} | = | \vec{b} |

.

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>EXERCISE>Q 15

If $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}, \vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}$ , find $λ$ such that $\vec{a}$ is perpendicular to $λ \vec{b} + \vec{c}$ .

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>EXERCISE>Q 16

If $\vec{p} = 5 \hat{i} + λ \hat{j} - 3 \hat{k}$ and $\vec{q} = \hat{i} + 3 \hat{j} - 5 \hat{k}$ then find the value of $λ$ , so that $\vec{p} + \vec{q}$ and $\vec{p} - \vec{q}$ are perpendicular vectors.