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…

Year	2026	2025	2024	Total
Questions	15	17	14	46

Q17 jee_main_2024_27_jan_morning Scalar Triple Product

Let

a= i+2 j+ k

,

b=3( i- j+ k)

. Let

c

be the vector such that

a\times c= b

and

a\cdot c=3

Then

a\cdot(( c\times b)- b- c)

equal to:

A. $32$ $32$
B. $24$ $24$
C. $20$ $20$
D. $36$ $36$

Solution

Related Formula

x \cdot ( y \times z) = [ x y z] = ( x \times y) \cdot z

Core Logic

The expression to evaluate is:

E = a \cdot (( c \times b) - b - c)

Distributing the dot product over the terms gives:

E = a \cdot ( c \times b) - a \cdot b - a \cdot c

Step 1: Evaluating the Scalar Triple Product

The first term is a scalar triple product:

a \cdot ( c \times b) = ( a \times c) \cdot b

We are given that $a \times c = b$ $\vec{a} \times \vec{c} = \vec{b}$. Substitute this in:

( b) \cdot b = | b|²

Given $b = 3 i - 3 j + 3 k$ , its magnitude squared is:

| b|² = 3² + (-3)² + 3² = 9 + 9 + 9 = 27

Step 2: Evaluating the remaining Dot Products

For the second term, calculate $a \cdot b$ $\vec{a} \cdot \vec{b}$:

a = 1 i + 2 j + 1 k

b = 3 i - 3 j + 3 k

a \cdot b = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0

For the third term, we are explicitly given:

a \cdot c = 3

Step 3: Final Output Calculation

Substitute all individual values back into the expanded expression:

E = 27 - 0 - 3 = 24

Pattern Recognition

When asked to evaluate complex vector expressions containing $a \cdot ( \times )$ $\vec{a} \cdot (\dots \times \dots)$, immediately distribute and convert them into Scalar Triple Products $[ a b c]$ $[\vec{a} \ \vec{b} \ \vec{c}]$. Cyclic permutations and given cross-product relationships will rapidly collapse the expression into simple magnitudes.

Chapter Mix

Class 12 Maths: Vector Algebra

Q21 jee_main_2024_27_jan_morning Dot Product

The least positive integral value of

α

, for which the angle between the vectors

α i-2 j+2 k

and

α i+2α j-2 k

is acute, is:

Numerical Answer. Answer: 5 to 5

Solution

Related Formula

θ = a \cdot b| a|| b|

For an acute angle, $θ > 0$ $\cos \theta > 0$, which strictly means $a \cdot b > 0$ $\vec{a} \cdot \vec{b} > 0$.

Core Logic

Given vectors $A = α i - 2 j + 2 k$ and $B = α i + 2α j - 2 k$ . For the angle to be acute, their dot product must be strictly positive:

A \cdot B > 0

(α)(α) + (-2)(2α) + (2)(-2) > 0

α² - 4α - 4 > 0

Step 1: Solving the Inequality

Complete the square to find critical points:

α² - 4α + 4 > 8

(α - 2)² > 8

Extracting roots gives:

α - 2 > 2\sqrt(2) or α - 2 < -2\sqrt(2)

α > 2 + 2\sqrt(2) or α < 2 - 2\sqrt(2)

Step 2: Finding Least Positive Integer

Approximate the boundary values. Since $\sqrt(2) \approx 1.414$ $\sqrt{2} \approx 1.414$: $2 + 2(1.414) = 4.828$ $2 + 2(1.414) = 4.828$

The ranges are $α in (-\infty, -0.828) (4.828, \infty)$ . We need the least positive integral value of $α$ $\alpha$. Looking at the interval $(4.828, \infty)$ $(4.828, \infty)$, the smallest integer present is $5$ $5$.

Pattern Recognition

Acute angle translates directly to a positive dot product. Set up the quadratic inequality, compute numerical bounds of irrational roots, and select the immediate next integer.

Chapter Mix

Class 12 Maths: Vector Algebra Class 11 Maths: Linear Inequalities

Q8 jee_main_2024_29_jan_morning Collinear Vectors

Let

a

,

b

and

c

be three non-zero vectors such that

b

and

c

are non-collinear. If

a+5 b

is collinear with

c

,

b+6 c

is collinear with

a

and

a+α b+β c=0

, then

α+β

is equal to

A. $35$ $35$
B. $30$ $30$
C. $-30$ $-30$
D. $-25$ $-25$

Solution

Related Formula

If X and Y are collinear, then X = λ Y for some scalar λ.

Linear Independence: If $u$ $\vec{u}$ and $v$ $\vec{v}$ are non-collinear, then $x u + y v = 0 x = 0 and y = 0$ .

Core Logic

Based on the problem statement:

$a + 5 b$ $\vec{a} + 5\vec{b}$ is collinear with $c$ $\vec{c}$:

$* a + 5 b = λ c (1)$

$b + 6 c$ $\vec{b} + 6\vec{c}$ is collinear with $a$ $\vec{a}$:

$* b + 6 c = μ a (2)$

Step 1: Eliminate Vector a

From equation (1), isolate $a$ $\vec{a}$:

a = λ c - 5 b

Substitute this into equation (2):

b + 6 c = μ(λ c - 5 b)

b + 6 c = μλ c - 5μ b

Rearrange to group coefficients for $b$ $\vec{b}$ and $c$ $\vec{c}$:

(1 + 5μ) b + (6 - μλ) c = 0

Step 2: Determine Scalars

Since vectors $b$ $\vec{b}$ and $c$ $\vec{c}$ are given as non-collinear, their linear combination resolving to zero dictates that their scalar coefficients must both individually be zero.

1 + 5μ = 0 \Rightarrow μ = -(1)/(5)

6 - μλ = 0 \Rightarrow 6 - (-(1)/(5))λ = 0 \Rightarrow 6 + (λ)/(5) = 0

(λ)/(5) = -6 \Rightarrow λ = -30

Step 3: Match the Final Equation

Substitute $λ = -30$ $\lambda = -30$ back into equation (1):

a + 5 b = -30 c

a + 5 b + 30 c = 0

The problem provides the structure $a + α b + β c = 0$ . Comparing the two yields:

α = 5, β = 30

Thus:

α + β = 5 + 30 = 35

Pattern Recognition

Double-collinearity equations should always be mapped out by eliminating the "third" vector ( $a$ $\vec{a}$ in this case) and funneling everything into the two known non-collinear vectors. The resulting zero-equation perfectly exposes the scalar unknowns.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Q13 jee_main_2024_29_jan_morning Vector Angle Bisector

Let O be the origin and the position vector of A and B be

2 i+2 j+ k

and

2 i+4 j+4 k

respectively. If the internal bisector of

∠ AOB

meets the line AB at C, then the length of OC is

A. $(2)/(3)\sqrt(31)$ $\frac{2}{3}\sqrt{31}$
B. $(2)/(3)\sqrt(34)$ $\frac{2}{3}\sqrt{34}$
C. $(3)/(4)\sqrt(34)$ $\frac{3}{4}\sqrt{34}$
D. $(3)/(2)\sqrt(31)$ $\frac{3}{2}\sqrt{31}$

Solution

Related Formula

Internal Angle Bisector Theorem: (AC)/(CB) = | OA|| OB|

Section Formula: OC = m OB + n OAm+n

Core Logic

Find the magnitudes of the position vectors $OA$ $\vec{OA}$ and $OB$ $\vec{OB}$:

| OA| = \sqrt(2² + 2² + 1²) = \sqrt(4+4+1) = \sqrt(9) = 3

| OB| = \sqrt(2² + 4² + 4²) = \sqrt(4+16+16) = \sqrt(36) = 6

According to the internal angle bisector theorem in $Δ AOB$ $\Delta AOB$, the point C divides the segment AB in the ratio of the adjacent sides:

(AC)/(CB) = | OA|| OB| = (3)/(6) = (1)/(2)

Step 1: Apply Section Formula

Using the section formula to find the position vector of C, dividing AB internally in ratio $m:n = 1:2$ $m:n = 1:2$:

OC = 1( OB) + 2( OA)1 + 2

OC = 1(2 i+4 j+4 k) + 2(2 i+2 j+ k)3

OC = (2+4) i + (4+4) j + (4+2) k3

OC = 6 i + 8 j + 6 k3 = 2 i + (8)/(3) j + 2 k

Step 2: Compute Length of OC

Now, find the magnitude (length) of the vector $OC$ $\vec{OC}$:

| OC| = \sqrt(2² + ((8)/(3))² + 2²)

= \sqrt(4 + (64)/(9) + 4) = \sqrt(8 + (64)/(9))

= \sqrt((72 + 64)/(9)) = \sqrt((136)/(9))

= \sqrt(4 \times 34)3 = 2\sqrt(34)3

Pattern Recognition

Vector angle bisector questions invariably test the geometric property that the bisector divides the opposite side in the ratio of the side lengths. Combine this directly with the 3D coordinate section formula.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Q4 jee_main_2024_30_january_evening Cross Product

Let

a = i +α j +β k,α ,β in R

.Let a vector

b

be such that the angle between

a

and

b

is

(π)/(4)

and

| b |² = 6

If

a\cdot b = 3\sqrt(2)

, then the value of

(α² + β²)| a\times b |²

is equal to

A. $90$ $90$
B. $75$ $75$
C. $95$ $95$
D. $85$ $85$

Solution

Related Formula

a\cdot b = | a|| b| θ

| a \times b|² = | a|² | b|² ² θ

| a|² = aₓ² + ay² + az²

Core Logic

Given $| b|² = 6$ $|\vec{b}|^2 = 6$, angle $θ = (π)/(4)$ $\theta = \frac{\pi}{4}$, and $a\cdot b = 3\sqrt(2)$ $\vec{a}\cdot\vec{b} = 3\sqrt{2}$.

| a|| b| θ = 3\sqrt(2)

Squaring both sides:

| a|² | b|² ² θ = 18

| a|² (6) ( 1\sqrt(2))² = 18

| a|² (6) ((1)/(2)) = 18 \Rightarrow 3| a|² = 18 \Rightarrow | a|² = 6

Step 1: Finding alpha and beta relation

We have $a = i + α j + β k$ .

| a|² = 1² + α² + β²

1 + α² + β² = 6 \Rightarrow α² + β² = 5

Step 2: Evaluating the Target Expression

We need to find the value of $(α² + β²)| a \times b|²$ .

| a \times b|² = | a|² | b|² ² θ

| a \times b|² = (6) (6) ²((π)/(4)) = 36 ((1)/(2)) = 18

Thus, the required value is:

(α² + β²)| a \times b|² = (5)(18) = 90

Pattern Recognition

Vector magnitude and dot product give direct length values. $| a\times b|² + ( a\cdot b)² = | a|²| b|²$ (Lagrange's Identity) skips $²θ$ $\sin^2\theta$ calculations completely.

Chapter Mix

Class 12 Maths: 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 8

Solution

Related Formula

Core Logic

Step 1: Evaluating the Scalar Triple Product

Step 2: Evaluating the remaining Dot Products

Step 3: Final Output Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solving the Inequality

Step 2: Finding Least Positive Integer

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Eliminate Vector a

Step 2: Determine Scalars

Step 3: Match the Final Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Section Formula

Step 2: Compute Length of OC

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding alpha and beta relation

Step 2: Evaluating the Target Expression

Pattern Recognition

Chapter Mix

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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: