Given four points P 1 , P 2 , P 3 and P 4 on the coordinate plane with origin O which satisfy the condition O P → n - 1 + O P → n + 1 = 3 2 O P → n , n = 2 , 3 . (b) If P 1 , P 2 , P 3 lie on the circle x 2 + y 2 = 1 , then prove that P 4 lies on this circle.

(b) Using parametric coordinates of circle x 2 + y 2 = 1 , let coordinates of P 1 , P 3 be cos α , sin α , cos β , sin β . For n = 2 , O P 1 → + O P 3 → = 3 2 O P 2 → i.e. O P 2 → = 2 3 cos α i ^ + sin α j ^ + cos β i ^ + sin β j ^ O P 2 → = 2 3 cos α + cos β i ^ + sin α + sin β j ^ i.e. coordinates of P 2 are 2 cos α + cos β 3 , 2 sin α + sin β 3 . Since it lies on the given circle, so 4 9 cos α + cos β 2 + sin α + sin β 2 = 1 ⇒ 2 + 2 cos α - β = 9 4 ⇒ cos α - β = 1 8 Now, for n = 3 , O P 2 → + O P 4 → = 3 2 O P 3 → O P 4 → = 3 2 O P 3 → - 2 3 O P 1 → + O P 3 → = 5 6 O P 3 → - 2 3 O P 1 → = 5 6 cos α - 2 3 cos β i ^ + 5 6 sin α - 2 3 sin β j ^ Now, O P 4 → 2 = 5 6 cos α - 2 3 cos β 2 + 5 6 sin α - 2 3 sin β 2 = 25 36 cos 2 α + 4 9 cos 2 β - 10 9 cos α cos β + 25 36 sin 2 α + 4 9 sin 2 β - 10 9 sin α sin β = 25 36 + 4 9 - 10 9 cos α - β = 25 36 + 4 9 - 10 9 · 1 8 = 1 ⇒ P 4 lies on x 2 + y 2 = 1

Solve the simultaneous vector equations for the vectors x → and y → . x → + c → × y → = a → and y → + c → × x → = b → where c → is a non zero vector.

Given: x → + c → × y → = a → … … i y → + c → × x → = b → … . . ii ⇒ y → = b → - c → × x → Putting in i , x → + c → × b → - c → × c → × x → = a → ⇒ x → - c → · x → c → + c → · c → x → = a → - c → × b → ⇒ 1 + c → 2 x → = a → - c → × b → + c → · x → c → . . . ( iii ) Taking dot product with c → in equation i , we get x → · c → = a → · c → ∵ c → × y → ⊥ c → ⇒ c → × y → · c → = 0 Putting in iii , x → = a → + a → · c → c → + b → × c → 1 + c → 2 So, ⇒ y → = b → - c → × a → + a → · c → c → + b → × c → 1 + c → 2 = b → - c → × a → + a → · c → c → × c → + c → × b → × c → 1 + c → 2 = b → - c → × a → + c → · c → b → - c → · b → c → 1 + c → 2 = b → 1 + c → 2 + a → × c → - c → 2 b → + c → · b → c → 1 + c → 2 ⇒ y → = b → + a → × c → + c → · b → c → 1 + c → 2

Let a 1 - a 2 a 1 - b 2 a 1 - c 2 b 1 - a 2 b 1 - b 2 b 1 - c 2 c 1 - a 2 c 1 - b 2 c 1 - c 2 = 0 and if the vectors α → = i ^ + a j ^ + a 2 k ^ ; β → = i ^ + b j ^ + b 2 k ^ ; γ → = i ^ + c j ^ + c 2 k ^ are non-coplanar, show that the vectors α → 1 = i ^ + a 1 j ^ + a 1 2 k ^ ; β → 1 = i ^ + b 1 j ^ + b 1 2 k ^ and γ → 1 = i ^ + c 1 j ^ + c 1 2 k ^ are coplanar.

α → = i ^ + a j ^ + a 2 k ^ , β → = i ^ + b j ^ + b 2 k ^ , γ → = i ^ + c j ^ + c 2 k ^ Since α → , β → , γ → are non-coplanar, ∴ 1 a a 2 1 b b 2 1 c c 2 ≠ 0 ⇒ a - b b - c c - a ≠ 0 ⇒ a ≠ b ≠ c Given a 1 - a 2 a 1 - b 2 a 1 - c 2 b 1 - a 2 b 1 - b 2 b 1 - c 2 c 1 - a 2 c 1 - b 2 c 1 - c 2 = 0 R 1 → R 1 - R 2 & R 2 → R 2 - R 3 , we get ⇒ a 1 - b 1 b 1 - c 1 a 1 + b 1 - 2 a a 1 + b 1 - 2 b a 1 + b 1 - 2 c b 1 + c 1 - 2 a b 1 + c 1 - 2 b b 1 + c 1 - 2 c c 1 - a 2 c 1 - b 2 c 1 - c 2 = 0 R 1 → R 1 - R 2 , we get ⇒ a 1 - b 1 b 1 - c 1 a 1 - c 1 a 1 - c 1 a 1 - c 1 b 1 + c 1 - 2 a b 1 + c 1 - 2 b b 1 + c 1 - 2 c c 1 - a 2 c 1 - b 2 c 1 - c 2 = 0 ⇒ a 1 - b 1 b 1 - c 1 c 1 - a 1 1 1 1 b 1 + c 1 - 2 a b 1 + c 1 - 2 b b 1 + c 1 - 2 c c 1 - a 2 c 1 - b 2 c 1 - c 2 = 0 C 1 → C 1 - C 2 & C 2 → C 2 - C 3 , we get ⇒ a 1 - b 1 b 1 - c 1 c 1 - a 1 0 0 1 2 b - a 2 c - b b 1 + c 1 - 2 c a 2 - b 2 - 2 c 1 a - b b 2 - c 2 - 2 c 1 b - c c 1 - c 2 ⏟ Δ = 0 ⇒ a 1 - b 1 b 1 - c 1 c 1 - a 1 Δ = 0 ⇒ a 1 - b 1 b 1 - c 1 c 1 - c 1 = 0 ∵ Δ ≠ 0 ⇒ 1 a 1 a 1 2 1 b 1 b 1 2 1 c 1 c 1 2 = 0 ⇒ α → 1 , β → 1 , γ → 1 are coplanar

If x → × y → = a → , y → × z → = b → , x → · b → = γ , x → · y → = 1 and y → · z → = 1 , then find x → , y → & z → in terms of a → , b → & γ .

x → × y → = a → ⇒ b → × x → × y → = b → × a → ⇒ b → · y → x → - b → · x → y → = b → × a → . . . ( 1 ) y → × z → = b → ⇒ y → · y → × z → = y → · b → ⇒ 0 = y → · b → ∵ y → ⊥ y → × z → ⇒ y → · y → × z → = 0 Putting in ( 1 ) , y → = a → × b → γ y → × z → = b → ⇒ y → × y → × z → = y → × b → ⇒ y → · z → y → - y → 2 z → = y → × b → ⇒ z → = y → + b → × y → y → 2 ⇒ z → = a → × b → γ + b → × a → × b → γ a → × b → γ 2 Now, x → × y → = a → ⇒ a → · y → = 0 ∵ y → ⊥ x → × y → ⇒ y → · x → × y → = 0 . . . ( 2 ) and a → × b → × x → × y → = a → × b → × a → ⇒ a → × b → · y → x → - a → × b → · x → y → = a → × b → × a → ⇒ x → = a → b → x → y → + a → × b → × a → a → b → y → . . . ( 3 ) Using y → × z → = b → ⇒ x → × y → × z → = x → × b → ⇒ x → · z → y → - x → · y → z → = x → × b → Taking dot product with a → , a → · y → x → . z → - a → · z → = a → x → b → From ( 2 ) , a → · y → = 0 ⇒ a → · z → = a → b → x → ∵ a → x → b → = - a → b → x → Now, y → × z → = b → ⇒ a → × y → × z → = a → × b → ⇒ a → · z → y → - a → · y → z → = a → × b → ⇒ a → · z → = γ a → · y → = 0 & γ y → = a → × b → ⇒ a → b → x → = γ So, x → = γ · a → × b → γ + a → × b → × a → a → × b → · a → × b → γ = a → × b → γ - a → × a → × b → γ a → × b → γ 2

Find the value of λ such that a , b , c are all non-zero and - 4 i ^ + 5 j ^ a + 3 i ^ - 3 j ^ + k ^ b + i ^ + j ^ + 3 k ^ c = λ a i ^ + b j ^ + c k ^

- 4 i ^ + 5 j ^ a + 3 i ^ - 3 j ^ + k ^ b + i ^ + j ^ + 3 k ^ c = λ a i ^ + b j ^ + c k ^ ⇒ i ^ - 4 a + 3 b + c + j ^ 5 a - 3 b + c + k ^ b + 3 c = λ a i ^ + b j ^ + c k ^ Comparing the respective components, we get - 4 a + 3 b + c = λ a ⇒ 3 b + c = λ + 4 a . . . . 1 5 a - 3 b + c = λ b ⇒ 5 a = b λ + 3 - c … … 2 b + 3 c = λ c ⇒ b = λ - 3 c . . . . . 3 From 1 & 3 , 3 λ - 3 c + c = λ + 4 a ⇒ a = 3 λ - 8 λ + 4 c . . ( 4 ) From ( 2 ) & ( 3 ) , 5 a = λ - 3 λ + 3 c - c ⇒ a = λ 2 - 10 5 c . . . ( 5 ) Since, a , c ≠ 0 ⇒ 3 λ - 8 λ + 4 = λ 2 - 10 5 ⇒ 15 λ - 40 = λ 3 + 4 λ 2 - 40 - 10 λ ⇒ λ λ 2 + 4 λ - 25 = 0 ⇒ λ = 0 , λ 2 + 4 λ - 25 = 0 ⇒ λ = 0 , - 2 ± 29 But for λ = 0 , a = b = c = 0 also become solutions. So, λ = - 2 ± 29

Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 4: EXERCISE-4

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 4: EXERCISE-4

Attempt the free practice questions on Chapter 21: Vector Algebra, Exercise 4: EXERCISE-4 with hints and solutions to strengthen your understanding. Beta Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 4: EXERCISE-4 with Hints & Solutions

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 1

Let $A B C D$ be a tetrahedron such that the edges $A B, A C$ and $A D$ are mutually perpendicular. Let the area of triangles $A B C, A C D$ and $A D B$ be denoted by $x, y$ and $z$ sq. units respectively. Find the area of the triangle $B C D$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 5

Find out whether the following pairs of lines are parallel, non-parallel & intersecting, or non-parallel & non-intersecting:

(b) $\vec{r_{1}} = \hat{i} - \hat{j} + 3 \hat{k} + λ (\hat{i} - \hat{j} + \hat{k})$ , $\vec{r_{2}} = 2 \hat{i} + 4 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + \hat{j} + 3 \hat{k})$

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 6

Find out whether the following pairs of lines are parallel, non-parallel & intersecting, or non-parallel & non-intersecting.

(c) $\vec{r_{1}} = \hat{i} + \hat{k} + λ (\hat{i} + 3 \hat{j} + 4 \hat{k})$ , $\vec{r_{2}} = 2 \hat{i} + 3 \hat{j} + μ (4 \hat{i} - \hat{j} + \hat{k})$

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 32

Given four points $P_{1}, P_{2}, P_{3}$ and $P_{4}$ on the coordinate plane with origin $O$ which satisfy the condition ${\vec{O P}}_{n - 1} + {\vec{O P}}_{n + 1} = \frac{3}{2} {\vec{O P}}_{n}, n = 2, 3$ .
(b) If $P_{1}, P_{2}, P_{3}$ lie on the circle $x^{2} + y^{2} = 1$ , then prove that $P_{4}$ lies on this circle.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 36

Solve the simultaneous vector equations for the vectors $\vec{x}$ and $\vec{y} . \vec{x} + \vec{c} \times \vec{y} = \vec{a}$ and $\vec{y} + \vec{c} \times \vec{x} = \vec{b}$ where $\vec{c}$ is a non zero vector.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 37

Let $|\begin{matrix} {(a_{1} - a)}^{2} & {(a_{1} - b)}^{2} & {(a_{1} - c)}^{2} \\ {(b_{1} - a)}^{2} & {(b_{1} - b)}^{2} & {(b_{1} - c)}^{2} \\ {(c_{1} - a)}^{2} & {(c_{1} - b)}^{2} & {(c_{1} - c)}^{2} \end{matrix}| = 0$ and if the vectors $\vec{α} = \hat{i} + a \hat{j} + a^{2} \hat{k}; \vec{β} = \hat{i} + b \hat{j} + b^{2} \hat{k}$ ; $\vec{γ} = \hat{i} + c \hat{j} + c^{2} \hat{k}$ are non-coplanar, show that the vectors ${\vec{α}}_{1} = \hat{i} + a_{1} \hat{j} + a_{1}^{2} \hat{k}$ ; ${\vec{β}}_{1} = \hat{i} + b_{1} \hat{j} + b_{1}^{2} \hat{k}$ and ${\vec{γ}}_{1} = \hat{i} + c_{1} \hat{j} + c_{1}^{2} \hat{k}$ are coplanar.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 39

If $\vec{x} \times \vec{y} = \vec{a}, \vec{y} \times \vec{z} = \vec{b}, \vec{x} \cdot \vec{b} = γ, \vec{x} \cdot \vec{y} = 1$ and $\vec{y} \cdot \vec{z} = 1$ , then find $\vec{x}, \vec{y} & \vec{z}$ in terms of $\vec{a}, \vec{b} & γ$ .

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 40

Find the value of $λ$ such that $a, b, c$ are all non-zero and $(- 4 \hat{i} + 5 \hat{j}) a + (3 \hat{i} - 3 \hat{j} + \hat{k}) b + (\hat{i} + \hat{j} + 3 \hat{k}) c = λ (a \hat{i} + b \hat{j} + c \hat{k})$

Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 4: EXERCISE-4

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 4: EXERCISE-4

Questions from Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 4: EXERCISE-4 with Hints & Solutions

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