JEE Advanced 1993

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Animated Solution for Mathematics - Vector Algebra: Let $a, b, c$ be distinct non-negative numbers. If the vectors $a \hat{i} + a \hat{j} + c \hat{k}, \hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ lie in a plane, then $c$ is

Q: Can I use the cross product method instead?

You absolutely can! You could find $\vec{v}_1 \times \vec{v}_2$ and then take the dot product with $\vec{v}_3$. However, that is mathematically identical to the scalar triple product. The determinant method is simply a more organized, less error-prone way to perform the exact same operation.

Select Answer:

Visualized Solution

Defining the Vectors

Let the given vectors be:
$v_{1} = a \hat{i} + a \hat{j} + c \hat{k}$
$v_{2} = 1 \hat{i} + 0 \hat{j} + 1 \hat{k}$
$v_{3} = c \hat{i} + c \hat{j} + b \hat{k}$

The Coplanarity Condition

The problem states these vectors lie in the same plane.
This means they are coplanar.

The Scalar Triple Product

For three coplanar vectors, the volume of the parallelepiped formed by them is zero.
Therefore, their Scalar Triple Product is zero:
$[v_{1} v_{2} v_{3}] = 0$

Setting up the Determinant

The Scalar Triple Product can be computed using a determinant.
$a 1 c a 0 c c 1 b = 0$

Choosing the Row for Expansion

To expand the determinant easily, look for zeros.
The second row ( $R_{2}$ ) has a zero in the middle.
We will expand along $R_{2}$ .

Expanding along $R_{2}$ (First Term)

The sign convention for $R_{2}$ is $(-, +, -)$ .
First term: $- 1 \cdot a c c b$

Expanding along $R_{2}$ (Second Term)

Second term: $+ 0 \cdot a c c b = 0$
This term vanishes completely!

Expanding along $R_{2}$ (Third Term)

Third term: $- 1 \cdot a c a c$
Notice the identical columns in this minor!

Evaluating the Minors

The equation becomes:
$- 1 (ab - c^{2}) - 1 (a c - a c) = 0$
The second minor evaluates to zero because $a c - a c = 0$ .

Simplifying the Equation

We are left with:
$- (ab - c^{2}) = 0$
Multiplying by $- 1$ :
$ab - c^{2} = 0$

The Final Relationship

Rearranging the terms:
$c^{2} = ab$
Since $a, b, c$ are non-negative, taking the square root gives:
$c = ab$

The Way Forward

The expression $c = ab$ is the standard formula for the Geometric Mean.
Therefore, $c$ is the Geometric Mean of $a$ and $b$ .
Final Answer: Option 2.

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The Sigma Insight: Scalar Triple Product

Solution Diagram

Analyzing the Setup

To determine the condition for the vectors

v_{1} = a \hat{i} + a \hat{j} + c \hat{k}

v_{2} = 1 \hat{i} + 0 \hat{j} + 1 \hat{k}

, and

v_{3} = c \hat{i} + c \hat{j} + b \hat{k}

to be coplanar, we must recognize that they occupy a single 2D plane.

In vector algebra, this implies that the volume of the parallelepiped formed by these three vectors is zero. Mathematically, this is expressed through the Scalar Triple Product.

The Master Equation

The condition for coplanarity is defined by the determinant of the matrix formed by the components of the vectors:

a 1 c a 0 c c 1 b = 0

To solve this efficiently, we expand along the second row, which contains a zero. This strategic choice simplifies the calculation significantly.

The Algebraic Expansion

Applying the cofactor expansion along the second row (using the sign convention

-, +, -

), we get:

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

Evaluating the first minor:

- (ab - c^{2})

Evaluating the second minor, we observe that the columns are identical (

a

and

c

in both columns), which results in zero:

- (a c - a c) = 0

Final Calculation

Combining these results, the equation simplifies to:

- (ab - c^{2}) = 0

ab - c^{2} = 0

c^{2} = ab

Given that

a, b,

and

c

are non-negative, we conclude that $c = ab$ . This confirms that

c

is the Geometric Mean of

a

and

b

Common Traps & FAQs

Wait, why do we set the scalar triple product to zero? Isn't it just a calculation?

Think of the scalar triple product as the volume of a parallelepiped. If the vectors are coplanar, they are trapped on a 2D plane. A 2D shape has zero height in 3D space, so the volume must be zero. It is a beautiful geometric constraint that turns a physical reality into a simple algebraic equation!

Why did we choose the second row to expand the determinant?

Efficiency is the mark of a JEE topper! The second row contains a zero. Expanding along a row with a zero eliminates one of the minor calculations entirely, saving you time and reducing the chance of a silly arithmetic error. Always scan your determinant for zeros before you start expanding.

Can I use the cross product method instead?

You absolutely can! You could find

v_{1} \times v_{2}

and then take the dot product with

v_{3}

. However, that is mathematically identical to the scalar triple product. The determinant method is simply a more organized, less error-prone way to perform the exact same operation.

Animated Solution for Mathematics - Vector Algebra: Let $a, b, c$ be distinct non-negative numbers. If the vectors $a \hat{i} + a \hat{j} + c \hat{k}, \hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ lie in a plane, then $c$ is

Select Answer:

Visualized Solution

Defining the Vectors

The Coplanarity Condition

The Scalar Triple Product

Setting up the Determinant

Choosing the Row for Expansion

Expanding along $R_{2}$ (First Term)

Expanding along $R_{2}$ (Second Term)

Expanding along $R_{2}$ (Third Term)

Evaluating the Minors

Simplifying the Equation

The Final Relationship

The Way Forward

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Master Equation

The Algebraic Expansion

Final Calculation

Similar Questions

Let $a, b$ and $c$ be distinct non-negative numbers. If the vectors $a \hat{i} + a \hat{j} + c \hat{k}$ , $\hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ lie in a plane, then $c$ is

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 $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} - \hat{j} + 2 \hat{k}$ and $c = x \hat{i} + (x - 2) \hat{j} - \hat{k}$ . If the vectors $c$ lies in the plane of $a$ and $b$ , then $x$ equals

Let $a = \hat{i} - \hat{j}$ , $b = \hat{j} - \hat{k}$ , $c = \hat{k} - \hat{i}$ . If $d$ is a unit vector such that $a \cdot d = 0 = [b c d]$ , then $d$ equals

If $a, b$ and $c$ are three non coplanar vectors, then $(a + b + c) \cdot [(a + b) \times (a + c)]$ equals

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 $a, b, c$ 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

For non-zero vectors $a, b, c$ , $∣ (a \times b) \cdot c ∣ = ∣ a ∣∣ b ∣∣ c ∣$ holds if and only if

Select Answer:

Visualized Solution

Defining the Vectors

The Coplanarity Condition

The Scalar Triple Product

Setting up the Determinant

Choosing the Row for Expansion

Expanding along .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​ (First Term)

Expanding along .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​ (Second Term)

Expanding along .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​ (Third Term)

Evaluating the Minors

Simplifying the Equation

The Final Relationship

The Way Forward

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Master Equation

The Algebraic Expansion

Final Calculation

Similar Questions

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Expanding along $R_{2}$ (First Term)

Expanding along $R_{2}$ (Second Term)

Expanding along $R_{2}$ (Third Term)

Let $a, b$ and $c$ be distinct non-negative numbers. If the vectors $a \hat{i} + a \hat{j} + c \hat{k}$ , $\hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ lie in a plane, then $c$ is

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Let $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} - \hat{j} + 2 \hat{k}$ and $c = x \hat{i} + (x - 2) \hat{j} - \hat{k}$ . If the vectors $c$ lies in the plane of $a$ and $b$ , then $x$ equals

Let $a = \hat{i} - \hat{j}$ , $b = \hat{j} - \hat{k}$ , $c = \hat{k} - \hat{i}$ . If $d$ is a unit vector such that $a \cdot d = 0 = [b c d]$ , then $d$ equals

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For non-zero vectors $a, b, c$ , $∣ (a \times b) \cdot c ∣ = ∣ a ∣∣ b ∣∣ c ∣$ holds if and only if