Prove Cauchy-Schwarz inequality: ( a → · b → ) 2 ≤ a → 2 · b → 2 .

Assume θ be the angle between vectors a → and b → . Then, a → · b → = | a → | | b → | cosθ ⇒ ( a → · b → ) 2 = | a → | 2 | b → | 2 cos 2 θ We know, cosθ ∈ [ - 1 , 1 ] Now, cos 2 θ ≤ 1 ⇒ a → 2 · b → 2 cos 2 θ ≤ | a | 2 | b | 2 ∴ ( a → · b → ) 2 ≤ a → 2 · b → 2 .

Prove Cauchy-Schwarz inequality: ( a → · b → ) 2 ≤ a → 2 · b → 2 .

Assume θ be the angle between vectors a → and b → . Then, a → · b → = | a → | | b → | cosθ ⇒ ( a → · b → ) 2 = | a → | 2 | b → | 2 cos 2 θ We know, cosθ ∈ [ - 1 , 1 ] Now, cos 2 θ ≤ 1 ⇒ a → 2 · b → 2 cos 2 θ ≤ | a | 2 | b | 2 ∴ ( a → · b → ) 2 ≤ a → 2 · b → 2 .

Two points P and Q are given in the rectangular Cartesian coordinates in the curve y = 2 x + 2 , such that O P → · i ^ = - 1 and O Q → · i ^ = 2 , where i ^ is a unit vector along the x -axis. Find the magnitude of O Q → - 4 O P → .

Assume P x 1 , y 1 and Q x 2 , y 2 be the two points on y = 2 x + 2 Projection of vector a → on vector b → = a → · b → | b → | a → · b → = | a → | | b → | cos θ O P → · i ^ is the projection of O P → on the x -axis ⇒ x 1 = – 1 Also, ( x 1 , y 1 ) lies on y = 2 x + 2 ⇒ y 1 = 2 x 1 + 2 ⇒ y 1 = 2 - 1 + 2 = 2 Also, O Q → · i ^ is the P rojection of O Q → on the x -axis. ⇒ x 2 = 2 ∵ O Q → . i ^ = 2 As x 2 , y 2 lies on y = 2 x + 2 ⇒ y 2 = 2 x 2 + 2 ⇒ y 2 = 16 Thus, O P → = x 1 i ^ + y 1 j ^ = - i ^ + 2 j ^ And O Q → = x 2 i ^ + y 2 j ^ = 2 i ^ + 16 j ^ ⇒ O Q → – 4 O P → = 6 i ^ + 8 j ^ ⇒ O Q → – 4 O P → = 36 + 64 = 10 .

O is the origin and A is a fixed point on the circle of radius a with centre O . The vector O A → is denoted by a → . A variable point lies on the tangent at A and O P → = r → . Show that a → . r → = a 2 . Hence, if P ( x , y ) and A ( x 1 , y 1 ) deduce the equation of tangent at A to this circle.

Assume A ( x 1 , y 1 ) in the x y – plane. ∴ O A → = a → = x 1 i ^ + y 1 j ^ OP → = r → = x i ^ + y j ^ Now A P = position vector of P - position vector of A AP → = ( r → – a → ) ∵ Point P lies on the tangent to the circle. ∴ O A is perpendicular to A P . ⇒ O A → · A P → = 0 ⇒ a → · ( r → - a → ) = 0 ⇒ a → · r → - a → . a → = 0 ⇒ a → · r → - a → 2 = 0 ⇒ a → · r → = a → 2 ⇒ x i ^ + y j ^ . x 1 i ^ + y 1 j ^ = a 2 ⇒ x x 1 + y y 1 = a 2 This is the equation of the tangent to the circle at the point A .

If a is real constant and A , B and C are variable angles and a 2 - 4 tan A + a tan B + a 2 - 4 tan C = 6 a , then find the least value of tan 2 A + tan 2 B + tan 2 C .

Given, a 2 - 4 tan A + a tan B + a 2 - 4 tan C = 6 a The given relation can be rewritten as, a 2 – 4 i ^ + a j ^ + a 2 – 4 k ^ . ( tanA i ^ + tanB j ^ + tanC k ^ ) = 6 a ( ∵ i ^ . i ^ = j ^ . j ^ = k ^ . k ^ = 1 ) ⇒ a 2 – 4 i ^ + a j ^ + a 2 – 4 k ^ . ( tanA i ^ + tanB j ^ + tanC k ^ ) = 6 a ⇒ a 2 – 4 i ^ + a j ^ + a 2 – 4 k ^ ( tanA i ^ + tanB j ^ + tanC k ^ ) cos θ = 6 a ⇒ a 2 – 4 + a 2 + ( a 2 + 4 ) tan 2 A + tan 2 + B + tan 2 C · cosθ = 6 a Now squaring both the sides ⇒ 3 a 2 ( tan 2 A + tan 2 B + tan 2 C ) cos 2 θ = 36 a 2 ⇒ tan 2 A + tan 2 B + tan 2 C = 12 sec 2 θ . . . . . ( i ) We know that the range of sec θ is ( - ∞ , - 1 ] ∪ [ 1 , ∞ ) ∴ sec 2 θ ≥ 1 ⇒ 12 sec 2 θ ≥ 12 . . . . . i i From Equations i and i i t a n 2 A + t a n 2 B + t a n 2 C ≥ 12 ∴ The l east value of t a n 2 A + t a n 2 B + tan 2 C = 12

MEDIUM

JEE Advanced

IMPORTANT

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Prove Cauchy-Schwarz inequality: $(\vec{a} \cdot \vec{b})^{2} \leq {|\vec{a}|}^{2} \cdot {|\vec{b}|}^{2}$ .

Important Questions on Product of Vectors

MEDIUM

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 2 - Product of Vectors>Exercise for Session 7>Q 165

Two points

P

and

Q

are given in the rectangular Cartesian coordinates in the curve

y = 2^{x + 2}

, such that

\vec{O P} \cdot \hat{i} = - 1

and

\vec{O Q} \cdot \hat{i} = 2,

where

\hat{i}

is a unit vector along the

x

-axis. Find the magnitude of

\vec{O Q} - 4 \vec{O P}

MEDIUM

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 2 - Product of Vectors>Exercise for Session 7>Q 166

O

is the origin and

A

is a fixed point on the circle of radius

a

with centre

O

. The vector

\vec{O A}

is denoted by

\vec{a}

. A variable point lies on the tangent at

A

and

\vec{O P} = \vec{r}

. Show that

\vec{a} . \vec{r} = a^{2}

. Hence, if

P (x, y)

and

A (x_{1}, y_{1})

deduce the equation of tangent at

A

to this circle.

MEDIUM

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 2 - Product of Vectors>Exercise for Session 7>Q 167

a

is real constant and

A, B

and

C

are variable angles and

\sqrt{a^{2} - 4} \tan A + a \tan B + \sqrt{a^{2} - 4} \tan C = 6 a

, then find the least value of

\tan^{2} A + \tan^{2} B + \tan^{2} C .

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 2 - Product of Vectors>Exercise for Session 7>Q 168

Given, the angle $A, B$ and $C$ of $∆ A B C$ . Find $∠ B A M$ , where $M$ is midpoint of $B C$ .

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 2 - Product of Vectors>Exercise for Session 7>Q 169

Find the perpendicular distance of

A (1, 4, - 2)

from the segment

B C

, where

B (2, 1, - 2)

and

C (0, - 5, 1)

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 2 - Product of Vectors>Exercise for Session 7>Q 170

Given angles

A, B

and

C

∆ A B C

. Let

M

be the midpoint of segment

A B

and let

D

be the foot of the bisector of

∠ C

. Find the ratio of

\frac{Area of ∆ C D M}{Area of ∆ A B C}

and also

\cos (∠ D C M)

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 2 - Product of Vectors>Exercise for Session 7>Q 171

In the

∆ A B C

a point

P

is taken on the side

A B

such that

A P : B P = 1 : 2

and a point

Q

is taken on the side

B C

such that

C Q : B Q = 2 : 1

, If

R

be the point of intersection of lines

A Q

and

C P

, using vector find the area of

∆ A B C

, if it is known that area of

∆ B R C

is one unit.

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 2 - Product of Vectors>Exercise for Session 7>Q 172

If one diagonal of a quadrilateral bisects the other, then it also bisects the quadrilateral.

Prove Cauchy-Schwarz inequality: (a→·b→)2≤a→2·b→2.

Important Questions on Product of Vectors

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Prove Cauchy-Schwarz inequality: $(\vec{a} \cdot \vec{b})^{2} \leq {|\vec{a}|}^{2} \cdot {|\vec{b}|}^{2}$ .