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Animated Solution for Mathematics - Vector Algebra: 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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Visualized Solution

Given Vectors

Let the given vectors be:
$u = a \hat{i} + a \hat{j} + c \hat{k}$
$v = \hat{i} + 0 \hat{j} + \hat{k}$
$w = c \hat{i} + c \hat{j} + b \hat{k}$

Coplanarity Condition

The vectors $u$ , $v$ , and $w$ are coplanar.
Geometrically, they lie on the same 2D plane in 3D space.

Scalar Triple Product

For three coplanar vectors, their Scalar Triple Product is zero.
$[u v w] = 0$
This means the determinant of their components equals zero.

Setting up the Determinant: Row 1

Row 1 contains the components of $u$ :
$a \dots \dots a \dots \dots c \dots \dots = 0$

Setting up the Determinant: Row 2

Row 2 contains the components of $v$ :
$a 1 \dots a 0 \dots c 1 \dots = 0$

Setting up the Determinant: Row 3

Row 3 contains the components of $w$ :
$a 1 c a 0 c c 1 b = 0$

Expanding the Determinant

We expand along Row 2 ( $R_{2}$ ) because it contains a zero, simplifying calculations.
The signs for $R_{2}$ expansion are: $-, +, -$

First Term of Expansion

Expanding the first element of $R_{2}$ :
$- 1 \cdot a c c b$
$- 1 (ab - c^{2})$

Remaining Terms of Expansion

The middle term is $0$ .
Expanding the last element of $R_{2}$ :
$- 1 \cdot a c a c$
Since two columns are identical, this determinant is $0$ .

Simplifying the Equation

Combining the terms:
$- 1 (ab - c^{2}) - 0 - 0 = 0$
$- (ab - c^{2}) = 0$
$ab - c^{2} = 0$

Final Conclusion

$c^{2} = ab$
This is the standard condition for a Geometric Progression.
Therefore, $c$ is the Geometric Mean of $a$ and $b$ .

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The Sigma Insight: Scalar Triple Product

Solution Diagram

Analyzing the Setup

We are given three vectors:

u = a \hat{i} + a \hat{j} + c \hat{k}

v = \hat{i} + \hat{k}

w = c \hat{i} + c \hat{j} + b \hat{k}

The problem states that these vectors are coplanar. Geometrically, this implies that the vectors lie on the same flat surface and cannot span any volume in 3D space.

The Scalar Triple Product

Whenever three vectors are coplanar, their scalar triple product must be exactly zero. Mathematically, this is expressed as:

[u v w] = 0

This product represents the volume of the parallelepiped formed by these vectors. To calculate this, we evaluate the determinant of their components:

a 1 c a 0 c c 1 b = 0

The Smart Expansion

To solve this determinant efficiently, we expand along the second row, which contains the most zeros:

1, 0, 1

. Following the sign convention for the second row

(-, +, -)

, we obtain:

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

The Revelation

Let us compute these smaller determinants. The first determinant is

(ab - c^{2})

, and the second determinant is

(a c - a c)

, which equals

0

Substituting these back into our equation:

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

This simplifies to:

ab - c^{2} = 0

c^{2} = ab

This is the classic condition for a Geometric Progression, indicating that

c

is the Geometric Mean of

a

and

b

. It is truly marvelous how the rigid, spatial nature of vectors leads us directly to the elegant world of sequences.

Common Traps & FAQs

Why is the scalar triple product zero?

Think of the scalar triple product as the volume of a parallelepiped. If three vectors lie in the same plane, they cannot 'lift' off the plane to create any height. Thus, the volume is zero. It is a beautiful geometric consequence of coplanarity.

Can I expand along any row?

Absolutely! You can expand along any row or column. However, choosing the row with the most zeros—like the second row here—is a strategic choice to minimize arithmetic errors and save time in the exam hall.

What if the vectors were linearly dependent?

Linear dependence is actually the algebraic manifestation of coplanarity. If three vectors are coplanar, one can be written as a linear combination of the other two, which is exactly what linear dependence means. They are two sides of the same coin!

Animated Solution for Mathematics - Vector Algebra: 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

Select Answer:

Visualized Solution

Given Vectors

Coplanarity Condition

Scalar Triple Product

Setting up the Determinant: Row 1

Setting up the Determinant: Row 2

Setting up the Determinant: Row 3

Expanding the Determinant

First Term of Expansion

Remaining Terms of Expansion

Simplifying the Equation

Final Conclusion

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Scalar Triple Product

The Smart Expansion

The Revelation

Similar Questions

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

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

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Select Answer:

Visualized Solution

Given Vectors

Coplanarity Condition

Scalar Triple Product

Setting up the Determinant: Row 1

Setting up the Determinant: Row 2

Setting up the Determinant: Row 3

Expanding the Determinant

First Term of Expansion

Remaining Terms of Expansion

Simplifying the Equation

Final Conclusion

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Scalar Triple Product

The Smart Expansion

The Revelation

Similar Questions

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