Let a = i + 2j + k and b = 2i + j - k. Let c be a unit vector in the plane of the vectors a and b and be perpendicular…

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More Vector Algebra Previous-Year Questions — Page 4

Q68 jee_main_2025_02_april_evening Properties of Vectors

Let

a = 2 i - 3 j + k

b = 3 i + 2 j + 5 k

and a vector

c

be such that

( a - c) \times b = -18 i - 3 j + 12 k

and

a \cdot c = 3

. If

b \times c = d

, then

| a \cdot d|

is equal to:

A. 18
B. 12
C. 9
D. 15

Solution

Related Formula

Vector Cross product distributes over subtraction: ( a - c) \times b = a \times b - c \times b

Scalar Triple Product cyclic identity: a \cdot ( b \times c) = ( a \times b) \cdot c

Antisymmetry: c \times b = - b \times c

Core Logic

Instead of solving for the individual coordinates of vector $c$ $\vec{c}$, we apply vector algebraic identities to compute the target scalar triple product directly.

Step 1: Expand and rewrite the cross product

Given $( a - c) \times b = -18 i - 3 j + 12 k$ :

a \times b - c \times b = -18 i - 3 j + 12 k

a \times b + b \times c = -18 i - 3 j + 12 k

b \times c = (-18 i - 3 j + 12 k) - ( a \times b) --- (1)

Step 2: Calculate a x b

Evaluate the cross product:

a \times b = vmatrix i & j & k 2 & -3 & 1 3 & 2 & 5 vmatrix

a \times b = i(-15 - 2) - j(10 - 3) + k(4 - (-9)) = -17 i - 7 j + 13 k

Step 3: Solve for the vector d

Substitute $a \times b$ $\vec{a} \times \vec{b}$ back into equation (1):

d = b \times c = (-18 i - 3 j + 12 k) - (-17 i - 7 j + 13 k)

d = - i + 4 j - k

Step 4: Compute the final dot product

Now compute the requested dot product:

a \cdot d = (2 i - 3 j + k) \cdot (- i + 4 j - k)

a \cdot d = 2(-1) + (-3)(4) + 1(-1) = -2 - 12 - 1 = -15

| a \cdot d | = 15

Pattern Recognition

Scalar triple product shortcut: Recognizing that $a \cdot d = a \cdot ( b \times c) = [ a b c ]$ allows you to find the scalar value through simple determinants and linear equations instead of solving for the vector components.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Q jee_main_2025_02_april_morning Vector Projections

a

is a nonzero vector such that its projections on the vectors

2 i - j + 2 k

i + 2 j - 2 k

and

k

are equal, then a unit vector along

a

is:

A. $1\sqrt(155)(-7 i + 9 j + 5 k)$
B. $1\sqrt(155)(-7 i + 9 j - 5 k)$
C. $1\sqrt(155)(7 i + 9 j + 5 k)$
D. $1\sqrt(155)(7 i + 9 j - 5 k)$

Solution

Related Formula

The projection of vector $a$ $\vec{a}$ onto vector $b$ $\vec{b}$ is:

Proj_ b a = a \cdot b| b|

Core Logic

Let $a = a₁ i + a₂ j + a₃ k$ . Set up equations equating the three projections to determine the ratios of components.

Step 1: Set up Projection Ratios

Let $b = 2 i- j+2 k | b| = 3$

Let $c = i+2 j-2 k | c| = 3$

Let $d = k | d| = 1$

Equating projections:

(2a₁ - a₂ + 2a₃)/(3) = (a₁ + 2a₂ - 2a₃)/(3) = (a₃)/(1)

Step 2: Solve System of Linear Equations

From the last equality:

2a₁ - a₂ + 2a₃ = 3a₃ 2a₁ - a₂ = a₃ (1)

a₁ + 2a₂ - 2a₃ = 3a₃ a₁ + 2a₂ = 5a₃ (2)

Multiply (1) by $2$ $2$ and add to (2):

4a₁ - 2a₂ + a₁ + 2a₂ = 2a₃ + 5a₃ 5a₁ = 7a₃ a₁ = (7)/(5)a₃

Substitute back to find $a₂$ $a_2$:

a₂ = 2a₁ - a₃ = 2((7)/(5)a₃) - a₃ = (9)/(5)a₃

Step 3: Normalize to Unit Vector

The vector components are in ratio $a₁ : a₂ : a₃ = (7)/(5) : (9)/(5) : 1 = 7 : 9 : 5$ .

Magnitude factor = \sqrt(7² + 9² + 5²) = \sqrt(49 + 81 + 25) = \sqrt(155)

Thus, the unit vector is:

a = 1\sqrt(155)(7 i + 9 j + 5 k)

Pattern Recognition

Looking at the options, only option (C) provides components with positive signs for all three unit basis elements matching our proportional derivation of $7:9:5$ $7:9:5$ directly.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Q74 jee_main_2025_03_april_evening Vector Products

Let

a = i + 2 j + k

b = 3 i - 3 j + 3 k

c = 2 i - j + 2 k

and

d

be a vector such that

b \times d = c \times d

and

a \cdot d = 4

. Then

|( a \times d)|²

is equal to

Numerical Answer. Answer: 128 to 128

Solution

Related Formula

Lagrange's Identity:

| u \times v|² = | u|² | v|² - ( u \cdot v)²

Also, if $u \times w = v \times w ( u - v) \times w = 0 w$ is parallel/collinear to $u - v$ $\vec{u} - \vec{v}$.

Core Logic

Rearranging the cross product:

b \times d - c \times d = 0 ( b - c) \times d = 0

Thus, $d = λ ( b - c)$ $\vec{d} = \lambda (\vec{b} - \vec{c})$.

Step 1: Finding vector $d$ $\vec{d}$

Calculate $( b - c)$ $(\vec{b} - \vec{c})$:

b - c = (3 i - 3 j + 3 k) - (2 i - j + 2 k) = i - 2 j + k

Therefore:

d = λ( i - 2 j + k)

Substitute into $a \cdot d = 4$ $\vec{a} \cdot \vec{d} = 4$:

λ( i + 2 j + k) \cdot ( i - 2 j + k) = 4

λ(1 - 4 + 1) = 4 -2λ = 4 λ = -2

Thus:

d = -2( i - 2 j + k) = -2 i + 4 j - 2 k

| d|² = (-2)² + 4² + (-2)² = 4 + 16 + 4 = 24

Step 2: Calculating $| a \times d|²$ $|\vec{a} \times \vec{d}|^2$ using Lagrange's Identity

Given $| a|² = 1² + 2² + 1² = 6$ $|\vec{a}|^2 = 1^2 + 2^2 + 1^2 = 6$:

| a \times d|² = | a|² | d|² - ( a \cdot d)²

| a \times d|² = 6 \times 24 - 4² = 144 - 16 = 128

Pattern Recognition

Applying Lagrange's Identity avoids computing the actual cross product determinants vector-by-vector. This is extremely efficient and reduces chances of sign errors.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Q69 jee_main_2025_07_april_morning Dot and Cross Products

Let the angle

θ, 0 < θ < (π)/(2)

between two unit vectors

a

and

b

⁻¹( \sqrt(65)9)

. If the vector

c = 3 a + 6 b + 9( a \times b)

, then the value of

9( c \cdot a) - 3( c \cdot b)

A. $31$ $31$
B. $27$ $27$
C. $29$ $29$
D. $24$ $24$

Solution

Related Formula

Properties of dot and cross vector setups:

a \cdot a = | a|² = 1

a \cdot b = | a|| b| θ = θ

a \cdot ( a \times b) = 0, b \cdot ( a \times b) = 0

Core Logic

Given $θ = \sqrt(65)9$ $\sin\theta = \frac{\sqrt{65}}{9}$. Since $θ$ $\theta$ lies in the first quadrant:

θ = \sqrt(1 - ²θ) = \sqrt(1 - (65)/(81)) = \sqrt((16)/(81)) = (4)/(9)

Now, let's take individual dot product equations using $c = 3 a + 6 b + 9( a \times b)$ :

Find $c \cdot a$ $\vec{c} \cdot \hat{a}$:

c \cdot a = 3( a \cdot a) + 6( b \cdot a) + 9(( a \times b) \cdot a)

c \cdot a = 3(1) + 6 θ + 0 = 3 + 6((4)/(9)) = 3 + (24)/(9) = (51)/(9)

Step 1: Compute Second Dot Product Term

Find $c \cdot b$ $\vec{c} \cdot \hat{b}$:

c \cdot b = 3( a \cdot b) + 6( b \cdot b) + 9(( a \times b) \cdot b)

c \cdot b = 3 θ + 6(1) + 0 = 3((4)/(9)) + 6 = (12)/(9) + 6 = (12 + 54)/(9) = (66)/(9) = (22)/(3)

Step 2: Substitute and Finalize Result

Evaluate the target expression layout:

Value = 9( c \cdot a) - 3( c \cdot b)

Value = 9((51)/(9)) - 3((22)/(3)) = 51 - 22 = 29

Pattern Recognition

Remember that a cross product vector $( a \times b)$ $(\hat{a} \times \hat{b})$ is orthogonal to both constituent vectors. Thus, taking their dot product evaluates to $0$ $0$ immediately, allowing you to ignore that entire component during computation steps.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Let a = i + 2 j + k$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and b = 2 i + j - k$\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$. Let c$\hat{c}$ be a unit vector in the plane of the vectors a$\vec{a}$ and b$\vec{b}$ and be perpendicular to a$\vec{a}$. Then such a vector c$\hat{c}$ is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Define Coplanar Structural Form

Step 2: Force Orthogonality Constraint

Step 3: Substitute and Normalize

Pattern Recognition

Chapter Mix

Reference Study Guides

More Vector Algebra Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Expand and rewrite the cross product

Step 2: Calculate a x b

Step 3: Solve for the vector d

Step 4: Compute the final dot product

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up Projection Ratios

Step 2: Solve System of Linear Equations

Step 3: Normalize to Unit Vector

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding vector d$\vec{d}$

Step 2: Calculating | a × d|²$|\vec{a} \times \vec{d}|^2$ using Lagrange's Identity

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Second Dot Product Term

Step 2: Substitute and Finalize Result

Pattern Recognition

Chapter Mix

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Welcome back

Let $a = i + 2 j + k$ and $b = 2 i + j - k$ . Let $c$ $\hat{c}$ be a unit vector in the plane of the vectors $a$ $\vec{a}$ and $b$ $\vec{b}$ and be perpendicular to $a$ $\vec{a}$. Then such a vector $c$ $\hat{c}$ is:

Step 1: Finding vector $d$ $\vec{d}$

Step 2: Calculating $| a \times d|²$ $|\vec{a} \times \vec{d}|^2$ using Lagrange's Identity