JEE Main 2023 (08 April Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Vector Algebra: Let the vectors $u_{1} = \hat{i} + \hat{j} + a \hat{k}$ , $u_{2} = \hat{i} + b \hat{j} + \hat{k}$ , and $u_{3} = c \hat{i} + \hat{j} + \hat{k}$ be coplanar. If the vectors $v_{1} = (a + b) \hat{i} + c \hat{j} + c \hat{k}$ , $v_{2} = a \hat{i} + (b + c) \hat{j} + a \hat{k}$ and $v_{3} = b \hat{i} + b \hat{j} + (c + a) \hat{k}$ are also coplanar, then $6 (a + b + c)$ is equal to

Q: Why does the scalar triple product equal zero for coplanar vectors?

The scalar triple product $\vec{A} \cdot (\vec{B} \times \vec{C})$ represents the volume of the parallelepiped formed by the three vectors. If they are coplanar, they lie on the same flat surface, meaning they span no 3D volume. Thus, the volume must be zero. It is a beautiful geometric interpretation of an algebraic condition.

Q: What if $abc$ wasn't zero? How would I solve for $a+b+c$?

If $abc$ were not zero, you would have a system of equations. You would need to look for another constraint or relationship between the variables. Always look for symmetry or patterns that allow you to isolate the variables. In this problem, the structure of the vectors was designed to lead to a clean cancellation.

Select Answer:

Progressive Hints

Hint 1

Visualize the geometry: when three vectors are coplanar, they lie on the same flat surface, meaning they span no 3D volume. This is the physical essence of the scalar triple product being zero. If you have vectors

u_{1}, u_{2}, u_{3}

, their coplanarity implies that the determinant formed by their components must be zero.

Hint 2

Don't rush into brute-force expansion of the second determinant. In JEE Advanced, time is your most precious resource, and complex algebra is a breeding ground for errors. Use the properties of determinants—specifically, row operations—to create zeros in the matrix. This is the secret weapon of a topper.

Hint 3

Start by setting up the determinant for the first set of vectors:

11 c 1 b 1 a 11 = 0

. Expand this to find the relationship

a + b + c = 2 + ab c

. Then, apply the row operation

R_{3} \to R_{3} - (R_{1} + R_{2})

to the second determinant to simplify it before expanding.

Visualized Solution

Coplanarity of Vectors $u$

Given vectors: $u_{1} = \hat{i} + \hat{j} + a \hat{k}$ , $u_{2} = \hat{i} + b \hat{j} + \hat{k}$ , $u_{3} = c \hat{i} + \hat{j} + \hat{k}$
Condition for coplanarity: $[u_{1} u_{2} u_{3}] = 0$

Setting up the First Determinant

The scalar triple product is the determinant of their components.
$11 c 1 b 1 a 11 = 0$

Expanding the Determinant

Expanding along $R_{1}$ :
$1 (b - 1) - 1 (1 - c) + a (1 - b c) = 0$
Simplifying: $b - 1 - 1 + c + a - ab c = 0$

Deriving the First Relation

Rearranging the terms:
$a + b + c - 2 - ab c = 0$
$a + b + c = 2 + ab c \dots (1)$

Coplanarity of Vectors $v$

Given vectors: $v_{1} = (a + b) \hat{i} + c \hat{j} + c \hat{k}$ , $v_{2} = a \hat{i} + (b + c) \hat{j} + a \hat{k}$ , $v_{3} = b \hat{i} + b \hat{j} + (c + a) \hat{k}$
Condition: $[v_{1} v_{2} v_{3}] = 0$

Setting up the Second Determinant

$a + b a b c b + c b c a c + a = 0$

Applying Row Operations

To simplify, apply $R_{3} \to R_{3} - (R_{1} + R_{2})$ :
New $R_{31} = b - (a + b + a) = - 2 a$
New $R_{32} = b - (c + b + c) = - 2 c$
New $R_{33} = (c + a) - (c + a) = 0$

The Simplified Determinant

The determinant becomes:
$a + b a - 2 a c b + c - 2 c c a 0 = 0$

Expanding the Simplified Determinant

Expanding along $R_{3}$ :
$- 2 a [a c - c (b + c)] - (- 2 c) [a (a + b) - a c] = 0$
$- 2 a (a c - b c - c^{2}) + 2 c (a^{2} + ab - a c) = 0$

Solving for $ab c$

Expanding the brackets:
$- 2 a^{2} c + 2 ab c + 2 a c^{2} + 2 a^{2} c + 2 ab c - 2 a c^{2} = 0$
Canceling terms: $4 ab c = 0 ⟹ ab c = 0$

Finding $a + b + c$

Substitute $ab c = 0$ into equation (1):
$a + b + c = 2 + 0$
$a + b + c = 2$

Final Calculation

We need to find the value of $6 (a + b + c)$ .
$6 \times 2 = 12$
Final Answer: $12$

00:00 / 00:00

The Sigma Insight: Scalar Triple Product

Solution Diagram

Analyzing the Geometric Constraint

In 3D space, vectors are considered coplanar if they lie on the same flat plane. Geometrically, this implies that the volume of the parallelepiped formed by these vectors is zero.

The scalar triple product, denoted by

[u_{1}, u_{2}, u_{3}]

, represents this volume. Therefore, for coplanar vectors, we must satisfy the condition:

11 c 1 b 1 a 11 = 0

Expanding this determinant along the first row, we obtain:

1 (b - 1) - 1 (1 - c) + a (1 - b c) = 0

b - 1 - 1 + c + a - ab c = 0

a + b + c = 2 + ab c

Evaluating the Second Set of Vectors

We are given a second set of coplanar vectors,

v_{1}, v_{2}, v_{3}

. Their coplanarity implies the following determinant must vanish:

a + b a b c b + c b c a c + a = 0

To simplify this, we apply the row operation

R_{3} \to R_{3} - (R_{1} + R_{2})

. This yields:

a + b a - 2 a c b + c - 2 c c a 0 = 0

Final Calculation

Expanding the simplified determinant, we find that the expression reduces to:

ab c = 0

Substituting this result back into our first master equation,

a + b + c = 2 + ab c

, we arrive at:

a + b + c = 2 + 0 = 2

The final required value is

6 \times (a + b + c)

, which is:

6 \times 2 = 12

Common Traps & FAQs

Why does the scalar triple product equal zero for coplanar vectors?

The scalar triple product

A \cdot (B \times C)

represents the volume of the parallelepiped formed by the three vectors. If they are coplanar, they lie on the same flat surface, meaning they span no 3D volume. Thus, the volume must be zero. It is a beautiful geometric interpretation of an algebraic condition.

Can I just expand the second determinant directly without row operations?

You could, but it would be a nightmare of algebraic terms. In JEE Advanced, time is your most precious resource. Row operations are not just a shortcut; they are a way to minimize the probability of calculation errors. Always look for ways to simplify the matrix before you start expanding.

What if

ab c

wasn't zero? How would I solve for

a + b + c

ab c

were not zero, you would have a system of equations. You would need to look for another constraint or relationship between the variables. Always look for symmetry or patterns that allow you to isolate the variables. In this problem, the structure of the vectors was designed to lead to a clean cancellation.

Select Answer:

Visualized Solution

Coplanarity of Vectors $u$

Setting up the First Determinant

Expanding the Determinant

Deriving the First Relation

Coplanarity of Vectors $v$

Setting up the Second Determinant

Applying Row Operations

The Simplified Determinant

Expanding the Simplified Determinant

Solving for $ab c$

Finding $a + b + c$

Final Calculation

The Sigma Insight: Scalar Triple Product

Analyzing the Geometric Constraint

Evaluating the Second Set of Vectors

Final Calculation

Similar Questions

If $a = 2 \hat{i} + \hat{j} + 3 \hat{k}, b = 3 \hat{i} + 3 \hat{j} + \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ are coplanar vectors and $a \cdot c = 5, b ⊥ c$ , then $122 (c_{1} + c_{2} + c_{3})$ is equal to

If vectors $a, b, c$ are coplanar, show that $a \cdot a b \cdot a c \cdot a a \cdot b b \cdot b c \cdot b a \cdot c b \cdot c c \cdot c = 0$ .

If four distinct points with position vectors $a, b, c$ and $d$ are coplanar, then $[a b c]$ is equal to

Let a, b and c be distinct positive numbers. If the vectors $a \hat{i} + a \hat{j} + c \hat{k}, \hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ are co-planar, then c is equal to:

Let the vectors $(2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k}, (1 + b) \hat{i} + 2 \hat{j} - b \hat{k}$ and $(2 + b) \hat{i} + 2 \hat{j} + (1 - b) \hat{k}, a, b, c \in R$ be co-planar. Then which of the following is true?

Let $a, b, c$ be three non-coplanar vectors and $p, q, r$ are vectors defined by the relations $p = \frac{b \times c}{[ a b c ]}, q = \frac{c \times a}{[ a b c ]}, r = \frac{a \times b}{[ a b c ]}$ then the value of the expression $(a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r$ is equal to

Let $a, b$ and $c$ be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be $a - b + c$ , $λ a - 3 b + 4 c$ , $- a + 2 b - 3 c$ and $2 a - 4 b + 6 c$ respectively. If AB, AC and AD are coplanar, then $λ$ is :

If the vectors $a = λ \hat{i} + μ \hat{j} + 4 \hat{k}, b = - 2 \hat{i} + 4 \hat{j} - 2 \hat{k}$ and $c = 2 \hat{i} + 3 \hat{j} + \hat{k}$ are coplanar and the projection of $a$ on the vector $b$ is $54$ units, then the sum of all possible values of $λ + μ$ is equal to

If the vectors $a \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k}$ ( $a \neq = b \neq = c \neq = 1$ ) are coplanar, then the value of $\frac{1}{( 1 - a )} + \frac{1}{( 1 - b )} + \frac{1}{( 1 - c )} = .........$

Let $a = \hat{i} + 2 \hat{j} + 4 \hat{k}$ , $b = \hat{i} + λ \hat{j} + 4 \hat{k}$ and $c = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$ be coplanar vectors. Then the non-zero vector $a \times c$ is :

Select Answer:

Visualized Solution

Coplanarity of Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u

Setting up the First Determinant

Expanding the Determinant

Deriving the First Relation

Coplanarity of Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v

Setting up the Second Determinant

Applying Row Operations

The Simplified Determinant

Expanding the Simplified Determinant

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } abc

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b+c

Final Calculation

The Sigma Insight: Scalar Triple Product

Analyzing the Geometric Constraint

Evaluating the Second Set of Vectors

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a=2i^+j^​+3k^,b=3i^+3j^​+k^ and c=c1​i^+c2​j^​+c3​k^ are coplanar vectors and a⋅c=5,b⊥c, then 122(c1​+c2​+c3​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If vectors a,b,c are coplanar, show that ​a⋅ab⋅ac⋅a​a⋅bb⋅bc⋅b​a⋅cb⋅cc⋅c​​=0.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If four distinct points with position vectors a,b,c and d are coplanar, then [abc] is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a, b and c be distinct positive numbers. If the vectors ai^+aj^​+ck^,i^+k^ and ci^+cj^​+bk^ are co-planar, then c is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the vectors (2+a+b)i^+(a+2b+c)j^​−(b+c)k^,(1+b)i^+2j^​−bk^ and (2+b)i^+2j^​+(1−b)k^,a,b,c∈R be co-planar. Then which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the vectors ai^+j^​+k^, i^+bj^​+k^ and i^+j^​+ck^ (a=b=c=1) are coplanar, then the value of (1−a)1​+(1−b)1​+(1−c)1​=.........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^+2j^​+4k^, b=i^+λj^​+4k^ and c=2i^+4j^​+(λ2−1)k^ be coplanar vectors. Then the non-zero vector a×c is :

Coplanarity of Vectors $u$

Coplanarity of Vectors $v$

Solving for $ab c$

Finding $a + b + c$

If $a = 2 \hat{i} + \hat{j} + 3 \hat{k}, b = 3 \hat{i} + 3 \hat{j} + \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ are coplanar vectors and $a \cdot c = 5, b ⊥ c$ , then $122 (c_{1} + c_{2} + c_{3})$ is equal to

If vectors $a, b, c$ are coplanar, show that $a \cdot a b \cdot a c \cdot a a \cdot b b \cdot b c \cdot b a \cdot c b \cdot c c \cdot c = 0$ .

If four distinct points with position vectors $a, b, c$ and $d$ are coplanar, then $[a b c]$ is equal to

Let a, b and c be distinct positive numbers. If the vectors $a \hat{i} + a \hat{j} + c \hat{k}, \hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ are co-planar, then c is equal to:

Let the vectors $(2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k}, (1 + b) \hat{i} + 2 \hat{j} - b \hat{k}$ and $(2 + b) \hat{i} + 2 \hat{j} + (1 - b) \hat{k}, a, b, c \in R$ be co-planar. Then which of the following is true?

If the vectors $a \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k}$ ( $a \neq = b \neq = c \neq = 1$ ) are coplanar, then the value of $\frac{1}{( 1 - a )} + \frac{1}{( 1 - b )} + \frac{1}{( 1 - c )} = .........$

Let $a = \hat{i} + 2 \hat{j} + 4 \hat{k}$ , $b = \hat{i} + λ \hat{j} + 4 \hat{k}$ and $c = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$ be coplanar vectors. Then the non-zero vector $a \times c$ is :