Problem 89 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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 \([2005]\) (A) the Geometric Mean of \(a\) and \(b\) (B) the Arithmetic Mean of \(a\) and \(b\) (C) equal to zero (D) the Harmonic Mean of \(a\) and \(b\)

Step-by-Step Solution

Answer

(A) the Geometric Mean of \(a\) and \(b\).

Step 1: Understand the Problem

We are given three vectors, and the problem states that these vectors lie in a plane. This means that the vectors are linearly dependent. Our task is to determine the relationship of the variable \(c\) with \(a\) and \(b\). The multiple-choice options suggest that \(c\) might be a mean of \(a\) and \(b\) or equal to zero.

Step 2: Identify Linear Dependence Condition

If three vectors are coplanar, their scalar triple product must be zero. The vectors given are \(\mathbf{v_1} = a \hat{i} + a \hat{j} + c \hat{k}\), \(\mathbf{v_2} = \hat{i} + \hat{k}\), and \(\mathbf{v_3} = c \hat{i} + c \hat{j} + b \hat{k}\). The condition for these vectors to be coplanar is given by the equation:\[(\mathbf{v_1} \times \mathbf{v_2}) \cdot \mathbf{v_3} = 0.\]

Step 3: Calculate Cross Product

First, we calculate the cross product \(\mathbf{v_1} \times \mathbf{v_2}\):\[\mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a & a & c \ 1 & 0 & 1 \end{vmatrix} = ((a \cdot 1 - c \cdot 0) \hat{i} - (a \cdot 1 - c \cdot 1) \hat{j} + (a \cdot 0 - a \cdot 1) \hat{k}).\]Simplifying, we get: \[\hat{i} (a) - \hat{j} (a - c) - \hat{k} a = a \hat{i} + (c - a) \hat{j} - a \hat{k}.\]

Step 4: Calculate Scalar Triple Product

Next, compute the dot product of \(\mathbf{v_1} \times \mathbf{v_2}\) with \(\mathbf{v_3}\):\[(a \hat{i} + (c-a) \hat{j} - a \hat{k}) \cdot (c \hat{i} + c \hat{j} + b \hat{k}) = a \cdot c + (c-a) \cdot c - a \cdot b.\]This simplifies to:\[a \cdot c + c^2 - ac - a \cdot b = c^2 - a \cdot b = 0.\]

Step 5: Solve for c

From the equation \(c^2 - a \cdot b = 0\), we rearrange to find\[c^2 = ab.\]Taking the square root of both sides, we find\[c = \sqrt{ab}.\]This shows that \(c\) is the geometric mean of \(a\) and \(b\).

Step 6: Conclusion

We have determined that for the three given vectors to lie in a plane, the condition \(c = \sqrt{ab}\) must hold true. Therefore, \(c\) is the geometric mean of \(a\) and \(b\).

Key Concepts

Scalar Triple Product

In vector algebra, the scalar triple product is a crucial concept for determining the coplanarity of three vectors. The scalar triple product combines three vectors, say \( \mathbf{u}, \mathbf{v}, \mathbf{w} \), using both the cross product and dot product as follows:\[(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}\]. This expression calculates a scalar, a single numeric value, which helps us understand whether these vectors lie within the same geometrical plane.

If the scalar triple product equals zero, the vectors are coplanar—meaning they lie in the same plane and are linearly dependent. This critical concept ties back to the condition of coplanarity, as seen in the exercise provided above: \((\mathbf{v_1} \times \mathbf{v_2}) \cdot \mathbf{v_3} = 0\). Here, the scalar triple product was used to confirm that the vectors share a common plane by yielding zero.

Understanding and being able to compute the scalar triple product are essential skills in vector algebra as they frequently appear in problems involving geometry and physics.

Geometric Mean

The geometric mean is a mathematical concept often encountered when dealing with non-negative numbers, especially in problems involving growth rates, geometry, or situations like our current vector problem. Unlike the arithmetic mean, which would simply involve adding two numbers and dividing, the geometric mean of two numbers \(a\) and \(b\) is calculated as \(\sqrt{ab}\). It gives a different kind of 'average', one that is particularly useful when we want to find a central tendency that considers the product of the data values.

In the original exercise, the concept of the geometric mean holds the key to solving the problem. We discovered that the condition for the vectors to lie in the same plane is \(c = \sqrt{ab}\), identifying \(c\) as the geometric mean of \(a\) and \(b\). This relationship emerges organically from the algebra involved in checking the coplanarity condition, and underscores the importance of well-known mathematical means in solving vector problems.

The geometric mean is particularly powerful when applied to problems involving proportional growth or configuration where multiplication of factors defines the conditions, as it elegantly captures such relationships.

Linearly Dependent Vectors

Linearly dependent vectors are vectors that are not independent of each other in their directionality. Essentially, it means one of the vectors can be written as a linear combination of the others. Whenever you hear that vectors are coplanar, it implies that they are linearly dependent.

For example, with vectors \(\mathbf{v_1}, \mathbf{v_2},\text{ and }\mathbf{v_3}\), if \((\mathbf{v_1} \times \mathbf{v_2}) \cdot \mathbf{v_3} = 0\), these vectors are linearly dependent. This condition indicates that they lie on a single plane, which means there is a linear relationship among them. This serves as a built-in check to mathematically establish their dependence on each other.

Understanding linear dependence is vital for solving complex geometric problems, like our exercise, because it reveals underlying connections between vectors. Once you determine vectors are linearly dependent, like the vectors given \((a \hat{i} + a \hat{j} + c \hat{k}, \hat{i} + \hat{k})\text{ and }(c \hat{i} + c \hat{j} + b \hat{k})\), you can use this to solve equations regarding their configuration, such as finding particular means or other relationships.