The position vectors of four points A , B , C , D are a → , b → , 2 a → + 3 b → , 2 a → - 3 b → respectively. Express the vectors A C → , B D → , C D → , A B → in terms of a → and b → .

Let the origin be O , then the position vectors of points A , B , C , D will be, O A → = a → = a → O B → = b → = b → O C → = c → = 2 a → + 3 b → O D → = d → = 2 a → - 3 b → So, A C → = O C → - O A → ⇒ A C → = 2 a → + 3 b → - a → ⇒ A C → = 2 - 1 a → + 3 b → ⇒ A C → = a → + 3 b → B D → = O D → - O B → ⇒ B D → = 2 a → - 3 b → - b → ⇒ B D → = 2 a → + - 3 - 1 b → ⇒ B D → = 2 a → - 4 b → C D → = O D → - O C → ⇒ C D → = 2 a → - 3 b → - 2 a → + 3 b → ⇒ C D → = 2 a → - 3 b → - 2 a → - 3 b → ⇒ C D → = 2 - 2 a → + - 3 - 3 b → ⇒ C D → = - 6 b → A B → = O B → - O A → ⇒ A B → = b → - a → Hence, the vectors A C → , B D → , C D → , A B → are a → + 3 b → , 2 a → - 4 b → , - 6 b → , b → - a → respectively.

If the components of a → are ( 1 , 2 ) and b → = i ^ - j ^ + a → , then what are the components of 2 a → - 3 b → ?

Given that, components of a → are 1 , 2 and b → = i ^ - j ^ + a → . So, vector a → = i ^ + 2 j ^ , b → = i ^ - j ^ + i ^ + 2 j ^ ⇒ b → = 2 i ^ + j ^ To find: 2 a → - 3 b → . So, 2 a → = 2 i ^ + 2 j ^ ⇒ 2 a → = 2 i ^ + 4 j ^ And, 3 b → = 3 2 i ^ + j ^ ⇒ 3 b → = 6 i ^ + 3 j ^ Therefore, 2 a → - 3 b → = 2 i ^ + 4 j ^ - 6 i ^ + 3 j ^ ⇒ 2 a → - 3 b → = 2 i ^ + 4 j ^ - 6 i ^ - 3 j ^ ⇒ 2 a → - 3 b → = - 4 i ^ + j ^ Hence, the components of vector 2 a → - 3 b → is - 4 , 1 .

The position vectors of the points A , B , C are 2 i ^ + j ^ - k ^ , 3 i ^ - 2 j ^ + k ^ and i ^ + 4 j ^ - 3 k ^ respectively, show that A , B , C are collinear.

Given that: position vectors of the points A , B , C are 2 i ^ + j ^ - k ^ , 3 i ^ - 2 j ^ + k ^ and i ^ + 4 j ^ - 3 k ^ respectively. Let the origin be O . Then position vectors will be, O A → = a → = 2 i ^ + j ^ - k ^ O B → = b → = 3 i ^ - 2 j ^ + k ^ O C → = c → = i ^ + 4 j ^ - 3 k ^ Vector A B → = Position vector of B - Position vector of A ⇒ A B → = O B → - O A → ⇒ A B → = 3 i ^ - 2 j ^ + k ^ - 2 i ^ + j ^ - k ^ ⇒ A B → = 3 i ^ - 2 j ^ + k ^ - 2 i ^ - j ^ + k ^ ⇒ A B → = 3 - 2 i ^ + - 2 - 1 j ^ + 1 + 1 k ^ ⇒ A B → = i ^ - 3 j ^ + 2 k ^ Vector A C → = Position vector of C - Position vector of A ⇒ A C → = O C → - O A → ⇒ A C → = i ^ + 4 j ^ - 3 k ^ - 2 i ^ + j ^ - k ^ ⇒ A C → = i ^ + 4 j ^ - 3 k ^ - 2 i ^ - j ^ + k ^ ⇒ A C → = 1 - 2 i ^ + 4 - 1 j ^ + - 3 + 1 k ^ ⇒ A C → = - i ^ + 3 j ^ - 2 k ^ Collinear vectors are the vectors which have same or parallel support irrespective of their magnitudes and directions. Here, A B → = - A C → So, vectors A B → and A C → have same support. They lie on the same line but on opposite direction. Vectors A B → and A C → have same initial point i.e. A . So, they are co-initial vectors. Hence, the points A , B , C are collinear.

E M B I B E

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 28

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IMPORTANT

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The vectors $\vec{a}$ and $\vec{b}$ are non-zero non-collinear. For what value of $x$ , the vectors $(x - 2) \vec{a} + \vec{b}$ and $(2 x + 1) \vec{a} - \vec{b}$ are collinear ?

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Important Questions on Vectors

EASY

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IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 29

The position vectors of four points

A, B, C, D

are

\vec{a}, \vec{b}, 2 \vec{a} + 3 \vec{b}, 2 \vec{a} - 3 \vec{b}

respectively. Express the vectors

\vec{A C}, \vec{B D}, \vec{C D}, \vec{A B}

in terms of

\vec{a}

and

\vec{b}

.

EASY

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IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 30

If the points

A, B, C

and

D

have coordinates

(0, 1), (1, 0), (1, 2)

and

(2, 1)

respectively, then prove that

\vec{A B} = \vec{C D}

.

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Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 30

If the coordinates of the points

A

and

B

in a plane are

(1, 1)

and

(2, 2)

respectively, then find the coordinates of point

C

such that

\vec{A B} = \vec{B C}

.

EASY

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IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 31

If the components of

\vec{a}

are

(1, 2)

and

\vec{b} = \hat{i} - \hat{j} + \vec{a}

, then what are the components of

2 \vec{a} - 3 \vec{b}

?

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Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 33

Show that the points

(0, - 1), (1, 0)

and

(3, 2)

are collinear.

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12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 33

Show that the points

A (2, 6, 3)

,

B (1, 2, 7)

and

C (3, 10, - 1)

are collinear.

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IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 33

The position vectors of the points

A, B, C

are

2 \hat{i} + \hat{j} - \hat{k}, 3 \hat{i} - 2 \hat{j} + \hat{k}

and

\hat{i} + 4 \hat{j} - 3 \hat{k}

respectively, show that

A, B, C

are collinear.

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IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 34

If $\vec{a}, \vec{b}$ are position vectors of points $A (1, - 1)$ and $B (- 3, m)$ , then find the value of $m$ so that the origin $O$ , and points $A$ and $B$ are not collinear.