Orthogonal Vectors and Scalar Product Calculation

Understanding the properties of vectors is crucial in mathematics. Two vectors are considered orthogonal if they are perpendicular to each other. This geometric condition has a direct algebraic implication: their dot product (also known as the scalar product) must be equal to zero.

Given Vectors for Orthogonality

We are provided with two vectors:

• Vector \(\overline{a}\): \(\overline{a} = 3\hat{i} - 2\hat{j} + \hat{k}\)
• Vector \(\overline{b}\): \(\overline{b} = 4\hat{i} + 3\hat{j} - \lambda\hat{k}\)

The problem states that these vectors \(\overline{a}\) and \(\overline{b}\) are orthogonal. Our goal is to find the value of \(\lambda\) that satisfies this condition of orthogonality.

Dot Product Application for Orthogonal Vectors

For any two vectors \(\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}\) and \(\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}\), their dot product is defined as:

\(\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z\)

Since \(\overline{a}\) and \(\overline{b}\) are orthogonal, their dot product must be zero. This is the fundamental condition for orthogonality:

\(\overline{a} \cdot \overline{b} = 0\)

Step-by-Step Calculation of \(\lambda\)

Let's calculate the dot product of the given vectors \(\overline{a}\) and \(\overline{b}\) using their components:

\(\overline{a} \cdot \overline{b} = (3\hat{i} - 2\hat{j} + \hat{k}) \cdot (4\hat{i} + 3\hat{j} - \lambda\hat{k})\)

Multiply the corresponding components (x with x, y with y, z with z) and sum them up:

\(\overline{a} \cdot \overline{b} = (3)(4) + (-2)(3) + (1)(-\lambda)\)

Now, perform the multiplications:

\(\overline{a} \cdot \overline{b} = 12 - 6 - \lambda\)

Simplify the expression:

\(\overline{a} \cdot \overline{b} = 6 - \lambda\)

As established, for the vectors to be orthogonal, their dot product must be equal to zero. So, we set the simplified expression to zero:

\(6 - \lambda = 0\)

To find the value of \(\lambda\), we isolate \(\lambda\) on one side of the equation:

\(\lambda = 6\)

Conclusion for \(\lambda\) Value

Therefore, the value of \(\lambda\) for which the vectors \(\overline{a} = 3\hat{i} - 2\hat{j} + \hat{k}\) and \(\overline{b} = 4\hat{i} + 3\hat{j} - \lambda\hat{k}\) are orthogonal is 6. This calculation aligns with the condition that the dot product of orthogonal vectors is zero, making \(\lambda = 6\) the correct solution.