If p → = 7 i ^ - 2 j ^ + 3 k ^ and q → = 3 i ^ + j ^ + 5 k ^ , then find the magnitude of p → – 2 q → .

Given, p → = 7 i ^ - 2 j ^ + 3 k ^ q → = 3 i ^ + j ^ + 5 k ^ To find, ; p → - 2 q → p → - 2 q → = 7 i ^ - 2 j ^ + 3 k ^ - 2 ( 3 i ^ + j ^ + 5 k ^ = i ^ - 4 j ^ - 7 k ^ = 1 2 + 4 2 + 7 2 = 66

If p → = 7 i ^ - 2 j ^ + 3 k ^ and q → = 3 i ^ + j ^ + 5 k ^ , then find the magnitude of p → – 2 q → .

Given, p → = 7 i ^ - 2 j ^ + 3 k ^ q → = 3 i ^ + j ^ + 5 k ^ To find, ; p → - 2 q → p → - 2 q → = 7 i ^ - 2 j ^ + 3 k ^ - 2 ( 3 i ^ + j ^ + 5 k ^ = i ^ - 4 j ^ - 7 k ^ = 1 2 + 4 2 + 7 2 = 66

Find a vector in the direction of 5 i ^ - j ^ + 2 k ^ , which has magnitude 8 units .

Let the given vector a → = 5 i ^ - j ^ + 2 k ^ , Now, we need to find a vector with magnitude 8 units in the direction of the given vector. We know that The unit vector in the direction of the given vector a → is given by a ^ = a → a → . Now, a → = 5 2 + - 1 2 + 2 2 = 25 + 1 + 4 = 30 Thus, a ^ = a → a → = 5 i ^ - j ^ + 2 k ^ 30 Hence, the vector in the direction of vector 5 i ^ - j ^ + 2 k ^ which has magnitude of 8 units is given by, = 8 a ^ = 8 5 i ^ - j ^ + 2 k ^ 30 = 40 30 i ^ - 8 30 j ^ + 16 30 k ^ .

If a → = i ^ + 2 j ^ + 2 k ^ and b → = 3 i ^ + 6 j ^ + 2 k ^ , then find a vector in the direction of a → and having a magnitude of | b → | .

Given, a → = i ^ + 2 j ^ + 2 k ^ , b → = 3 i ^ + 6 j ^ + 2 k ^ Unit vector, a ^ = 1 1 2 + 2 2 + 2 2 i ^ + 2 j ^ + 2 k ^ = 1 3 i ^ + 2 j ^ + 2 k ^ . Vector of magnitude b → along a → = b → a ^ = ( 3 ) 2 + ( 6 ) 2 + ( 2 ) 2 × 1 3 ( i ^ + 2 j ^ + 2 k ^ ) = 7 3 ( i ^ + 2 j ^ + 2 k ^ ) .

Find the position vector of a point P which divides the line joining two points A and B whose position vectors are i ^ + 2 j ^ - k ^ and - i ^ + j ^ + k ^ respectively in the ratio 2 : 1 i internally i i externally

Given, A = i ^ + 2 j ^ - k ^ , B = - i ^ + j ^ + k ^ We know that, the position vector of a point P dividing a line segment joining the points A and B whose position vectors are a → and b → respectively, in the ratio m : n externally and internally are given by m b → - n a → m - n and m b → + n a → m + n respectively. So, position vector of a point P which divides the line A B in the ratio of 2 : 1 externally is O P → = 2 - i ^ + j ^ + k ^ - 1 i ^ + 2 j ^ - k ^ 2 - 1 ⇒ O P → = - 2 i ^ + 2 j ^ + 2 k ^ - ( i ^ + 2 j ^ - k ^ ) 2 - 1 = - 2 i ^ + 2 j ^ + 2 k ^ - i ^ + 2 j ^ - k ^ = - 3 i ^ + 3 k ^ . And position vector of a point P which divides the line A B in the ratio 2 : 1 internally is O P → = 2 - i ^ + j ^ + k ^ + 1 · i ^ + 2 j ^ - k ^ 2 + 1 ⇒ O P → = - 1 3 i ^ + 4 3 j ^ + 1 3 k ^ .

If the position vector of one end of a line segment A B → is 2 i ^ + 3 j ^ - k ^ and the position vector of its middle point is 3 i ^ + j ^ + k ^ , then find the position vector of the other end.

Let A → = 2 i ^ + 3 j ^ - k ^ & B → = r → Middle point of A B → = 3 i ^ + j ^ + k ^ Now, 3 i ^ + j ^ + k ^ = 2 i ^ + 3 j ^ - k ^ + r → 2 ⇒ 6 i ^ + j ^ + k ^ = 2 i ^ + 3 j ^ - k ^ + r → ⇒ r → = 4 i ^ + 3 j ^ + 7 k ^ So, the point B → is 4 i ^ + 3 j ^ + 7 k ^ .

If the position vectors of the points A → , B → and C → are a → , b → and 3 a → – 2 b → respectively, then prove that the points A → , B → and C → are collinear.

Given: a → , b → and 3 a → – 2 b → are the position vectors of A → , B → and C → respectively. To prove: The given points are collinear. Proof: A B → = b → - a → , B C → = 3 a → - 3 b → , A C → = 2 a → - 2 b → A B → × B C → . A C → = b → - a → × 3 a → - 3 b → . ( 2 a → - 2 b → ) = - 3 a → - b → × a → - b → . 2 a → - 2 b → = 0 (As a → × a → = 0 → ) Hence, A → , B → , C → are collinear.

The position vectors of four points P , Q , R and S are 2 a → + 4 c → , 5 a → + 3 3 b → + 4 c → , - 2 3 b → + c → and 2 a → + c → respectively. Prove that P Q → is parallel to R S → .

Given: Position vectors of P , Q , R & S are 2 a → + 4 c → , 5 a → + 3 3 b → + 4 c → , - 2 3 b → + c → and 2 a → + c → respectively. To prove: P Q → is parallel to R S → P Q → = 3 a → + 3 3 b → R S → = 2 a → + 2 3 b → P Q → × R S → = 3 a → + 3 3 b → × ( 2 a → + 2 3 b → ) ⇒ P Q → × R S → = 6 a → + 3 b → × ( a → + 3 b → ) = 0 → Hence, P Q → ∥ R S → .

E M B I B E

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Exercise for Session 2>Q 13

EASY

JEE Advanced

IMPORTANT

Earn 100

If $\vec{p} = 7 \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\vec{q} = 3 \hat{i} + \hat{j} + 5 \hat{k}$ , then find the magnitude of $\vec{p} - 2 \vec{q}$ .

Important Questions on Vector Algebra

EASY

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Exercise for Session 2>Q 14

Find a vector in the direction of

5 \hat{i} - \hat{j} + 2 \hat{k}

, which has magnitude

8 units

.

EASY

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Exercise for Session 2>Q 15

If $\vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{b} = 3 \hat{i} + 6 \hat{j} + 2 \hat{k}$ , then find a vector in the direction of $\vec{a}$ and having a magnitude of $| \vec{b} |$ .

MEDIUM

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Exercise for Session 2>Q 16

Find the position vector of a point $P$ which divides the line joining two points $A$ and $B$ whose position vectors are $\hat{i} + 2 \hat{j} - \hat{k}$ and $- \hat{i} + \hat{j} + \hat{k}$ respectively in the ratio $2 : 1$

$(i)$ internally

(i i)

externally

MEDIUM

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Exercise for Session 2>Q 16

Find the position vector of a point $P$ which divides the line joining two points $A & B$ , whose position vectors are $\hat{i} + 2 \hat{j} - \hat{k} and - \hat{i} + \hat{j} + \hat{k}$ , respectively in the ratio of $2 : 1$ externally.