Show that the points A , B and C with position vectors 3 i ^ - 4 j ^ - 4 k ^ , 2 i ^ - j ^ + k ^ and i ^ - 3 j ^ - 5 k ^ respectively form the vertices of a right-angled triangle.

Given vertices of a triangle are A 3 i ^ - 4 j ^ - 4 k ^ , B 2 i ^ - j ^ + k ^ and C i ^ - 3 j ^ - 5 k ^ . To prove: The given points are the vertices of a right-angled triangle. Let the origin be O . Then the position vectors of the vertices will be, O A → = a → = 3 i ^ - 4 j ^ - 4 k ^ O B → = b → = 2 i ^ - j ^ + k ^ O C → = c → = i ^ - 3 j ^ - 5 k ^ Here, vector of sides of the triangle are, Vector of side A B , A B → = Position vector of B - Position vector of A ⇒ A B → = O B → - O A → ⇒ A B → = 2 i ^ - j ^ + k ^ - 3 i ^ - 4 j ^ - 4 k ^ ⇒ A B → = 2 i ^ - j ^ + k ^ - 3 i ^ + 4 j ^ + 4 k ^ ⇒ A B → = 2 - 3 i ^ + - 1 + 4 j ^ + 1 + 4 k ^ ⇒ A B → = - i ^ + 3 j ^ + 5 k ^ Vector of side B C , B C → = O C → - O B → ⇒ B C → = i ^ - 3 j ^ - 5 k ^ - 2 i ^ - j ^ + k ^ ⇒ B C → = i ^ - 3 j ^ - 5 k ^ - 2 i ^ + j ^ - k ^ ⇒ B C → = 1 - 2 i ^ + - 3 + 1 j ^ + - 5 - 1 k ^ ⇒ B C → = - i ^ - 2 j ^ - 6 k ^ Vector of side A C , A C → = O C → - O A → ⇒ A C → = i ^ - 3 j ^ - 5 k ^ - 3 i ^ - 4 j ^ - 4 k ^ ⇒ A C → = i ^ - 3 j ^ - 5 k ^ - 3 i ^ + 4 j ^ + 4 k ^ ⇒ A C → = 1 - 3 i ^ + - 3 + 4 j ^ + - 5 + 4 k ^ ⇒ A C → = - 2 i ^ + j - k ^ Therefore, Magnitude of A B → , A B → = - 1 2 + 3 2 + 5 2 ⇒ A B → = 1 + 9 + 25 ⇒ A B → = 35 Magnitude of B C → , B C → = - 1 2 + - 2 2 - 6 2 ⇒ B C → = 1 + 4 + 36 ⇒ B C → = 41 Magnitude of A C → , A C → = - 2 2 + 1 2 + - 1 2 ⇒ A C → = 4 + 1 + 1 ⇒ A C → = 6 Applying Pythagoras theorem, B C 2 = A B 2 + A C 2 ⇒ 41 2 = 35 2 + 6 2 ⇒ 41 = 41 Pythagoras theorem is verified. Hence, the points A , B and C are the vertices of a right-angled triangle.

If the position vectors of the vertices of a triangle A B C are i ^ + 2 j ^ + 3 k ^ , 2 i ^ + 3 j ^ + k ^ and 3 i ^ + j ^ + 2 k ^ , then prove that △ A B C is an equilateral triangle.

Given that, vertices of a triangle A B C are A i ^ + 2 j ^ + 3 k ^ , B 2 i ^ + 3 j ^ + k ^ and C 3 i ^ + j ^ + 2 k ^ . Let the origin be O . Then position vectors of the points will be, O A → = i ^ + 2 j ^ + 3 k ^ O B → = 2 i ^ + 3 j ^ + k ^ O C → = 3 i ^ + j ^ + 2 k ^ Therefore, Vector of side A B , A B → = O B → - O A → ⇒ A B → = 2 i ^ + 3 j ^ + k ^ - i ^ + 2 j ^ + 3 k ^ ⇒ A B → = 2 i ^ + 3 j ^ + k ^ - i ^ - 2 j ^ - 3 k ^ ⇒ A B → = 2 - 1 i ^ + 3 - 2 j ^ + 1 - 3 k ^ ⇒ A B → = i ^ + j ^ - 2 k ^ Vector of side B C , B C → = O C → - O B → ⇒ B C → = 3 i ^ + j ^ + 2 k ^ - 2 i ^ + 3 j ^ + k ^ ⇒ B C → = 3 i ^ + j ^ + 2 k ^ - 2 i ^ - 3 j ^ - k ^ ⇒ B C → = 3 - 2 i ^ + 1 - 3 j ^ + 2 - 1 k ^ ⇒ B C → = i ^ - 2 j ^ + k ^ Vector of side C A , C A → = O A → - O C → ⇒ C A → = i ^ + 2 j ^ + 3 k ^ - 3 i ^ + j ^ + 2 k ^ ⇒ C A → = i ^ + 2 j ^ + 3 k ^ - 3 i ^ - j ^ - 2 k ^ ⇒ C A → = 1 - 3 i ^ + 2 - 1 j ^ + 3 - 2 k ^ ⇒ C A → = - 2 i ^ + j ^ + k ^ Magnitude of vector A B → , A B → = 1 2 + 1 2 + - 2 2 ⇒ A B → = 1 + 1 + 4 ⇒ A B → = 6 Magnitude of vector B C → , B C → = 1 2 + - 2 2 + 1 2 ⇒ B C → = 1 + 4 + 1 ⇒ B C → = 6 Magnitude of vector C A → , C A → = - 2 2 + 1 2 + 1 2 ⇒ C A → = 4 + 1 + 1 ⇒ C A → = 6 Here, magnitude of all the three sides of the triangle are equal. Hence, the △ A B C is an equilateral triangle.

The position vectors a → , b → , c → of three given points satisfy the relation 4 a → - 9 b → + 5 c → = 0 → . Prove that the three points are collinear.

Let the a → , b → , c → are the position vectors of points A , B and C . To prove: The points A , B and C are collinear. Given that: 4 a → - 9 b → + 5 c → = 0 → So, 4 a → - 9 b → + 5 c → = 0 → ⇒ 9 b → = 4 a → + 5 c → ⇒ b → = 4 a → + 5 c → 9 ⇒ b → = 4 a → + 5 c → 4 + 5 On comparing the above equation with section formula for internal division r → = m b → + n a → m + n , we can say that point B divides the line A C internally in the ratio m : n = 4 : 5 . Here " + v e " sign indicates that B divides the line A C internally. Hence, points A , B , C lies on same line, thus the points are collinear.

E M B I B E

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

MEDIUM

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IMPORTANT

Earn 100

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.

Important Questions on Vectors

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IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 35

Using vectors, find the value of

λ

such that the points

(λ, - 10, 3), (1, - 1, 3)

and

(3, 5, 3)

are collinear.

MEDIUM

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IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 36

If

\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 4 \hat{i} - 2 \hat{j} + 3 \hat{k}

and

\vec{c} = \hat{i} - 2 \hat{j} + \hat{k}

, then find a vector of magnitude

6

units which is parallel to the vector

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

.

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IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 37

Show that the points

A

,

B

and

C

with position vectors

3 \hat{i} - 4 \hat{j} - 4 \hat{k}, 2 \hat{i} - \hat{j} + \hat{k}

and

\hat{i} - 3 \hat{j} - 5 \hat{k}

respectively form the vertices of a right-angled triangle.

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IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 37

If the vertices of a triangle are

A (2, - 1, 1)

,

B (1, - 3, - 5)

and

C (3, - 4, - 4)

, then prove by vector method that it is a right-angled triangle.

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12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 37

If the position vectors of the vertices of a triangle

A B C

are

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

,

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

and

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

, then prove that

△ A B C

is an equilateral triangle.

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 38

If $\vec{a}, \vec{b}$ are position vectors of $A, B$ with reference to origin $O$ . find the position vector of a point $C$ in $A B$ produced such that $2 A C = 3 A B$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 1 - Vectors>EXERCISE>Q 39

The position vectors $\vec{a}, \vec{b}, \vec{c}$ of three given points satisfy the relation $4 \vec{a} - 9 \vec{b} + 5 \vec{c} = \vec{0}$ . Prove that the three points are collinear.