Understanding Perpendicular Vectors and the Dot Product

The question asks us to find the value of \({\rm{\lambda}}\) such that the two given vectors, \({\rm{\vec a}} = 2{\rm{\hat i}} + 3{\rm{\hat j}} + 4{\rm{\hat k}}\) and \({\rm{\vec b}} = 3{\rm{\hat i}} + 2{\rm{\hat j}} - {\rm{\lambda \hat k}}\), are perpendicular to each other.

In vector algebra, two non-zero vectors are perpendicular if and only if their dot product (also known as the scalar product) is zero. The dot product of two vectors \({\rm{\vec a}} = a_x{\rm{\hat i}} + a_y{\rm{\hat j}} + a_z{\rm{\hat k}}\) and \({\rm{\vec b}} = b_x{\rm{\hat i}} + b_y{\rm{\hat j}} + b_z{\rm{\hat k}}\) is calculated as:

$${\rm{\vec a}} \cdot {\rm{\vec b}} = a_x b_x + a_y b_y + a_z b_z$$

For the vectors to be perpendicular, we must have \({\rm{\vec a}} \cdot {\rm{\vec b}} = 0\).

Calculating the Dot Product of the Given Perpendicular Vectors

Let's identify the components of the given vectors:

• For \({\rm{\vec a}} = 2{\rm{\hat i}} + 3{\rm{\hat j}} + 4{\rm{\hat k}}\): \(a_x = 2\), \(a_y = 3\), \(a_z = 4\)
• For \({\rm{\vec b}} = 3{\rm{\hat i}} + 2{\rm{\hat j}} - {\rm{\lambda \hat k}}\): \(b_x = 3\), \(b_y = 2\), \(b_z = -{\rm{\lambda}}\)

Now, we calculate the dot product \({\rm{\vec a}} \cdot {\rm{\vec b}}\) using the formula:

$${\rm{\vec a}} \cdot {\rm{\vec b}} = (2)(3) + (3)(2) + (4)(-{\rm{\lambda}})$$

$${\rm{\vec a}} \cdot {\rm{\vec b}} = 6 + 6 - 4{\rm{\lambda}}$$

$${\rm{\vec a}} \cdot {\rm{\vec b}} = 12 - 4{\rm{\lambda}}$$

Finding the Value of Lambda for Perpendicularity

Since the vectors are perpendicular, their dot product must be equal to zero:

$${\rm{\vec a}} \cdot {\rm{\vec b}} = 0$$

Substitute the calculated dot product into this equation:

$$12 - 4{\rm{\lambda}} = 0$$

Now, we solve for \({\rm{\lambda}}\):

Add \(4{\rm{\lambda}}\) to both sides of the equation:

$$12 = 4{\rm{\lambda}}$$

Divide both sides by 4:

$${\rm{\lambda}} = \frac{12}{4}$$

$${\rm{\lambda}} = 3$$

Thus, the value of \({\rm{\lambda}}\) that makes the vectors \({\rm{\vec a}}\) and \({\rm{\vec b}}\) perpendicular is 3.

Revision Table: Key Concepts for Perpendicular Vectors

Additional Information: Properties of Perpendicular Vectors and Dot Product

The dot product is a fundamental operation in vector algebra with several important properties, especially concerning perpendicular vectors.

• Geometric Interpretation: The dot product \({\rm{\vec a}} \cdot {\rm{\vec b}}\) is also equal to \(|{\rm{\vec a}}||{\rm{\vec b}}|\cos\theta\), where \(|{\rm{\vec a}}|\) and \(|{\rm{\vec b}}|\) are the magnitudes of the vectors and \(\theta\) is the angle between them.
• Perpendicularity and Cosine: If vectors are perpendicular, the angle \(\theta = 90^\circ\). Since \(\cos(90^\circ) = 0\), their dot product \(|{\rm{\vec a}}||{\rm{\vec b}}|\cos(90^\circ) = |{\rm{\vec a}}||{\rm{\vec b}}|(0) = 0\). This confirms the condition \({\rm{\vec a}} \cdot {\rm{\vec b}} = 0\) for perpendicular vectors.
• Parallel Vectors: If vectors are parallel, the angle \(\theta\) is \(0^\circ\) or \(180^\circ\). The dot product would be \(|{\rm{\vec a}}||{\rm{\vec b}}|\) or \(-|{\rm{\vec a}}||{\rm{\vec b}}|\) respectively, not zero (unless one vector is the zero vector).
• Self Dot Product: The dot product of a vector with itself is the square of its magnitude: \({\rm{\vec a}} \cdot {\rm{\vec a}} = |{\rm{\vec a}}|^2\).

Understanding the dot product and its relation to the angle between vectors is crucial for solving problems involving perpendicularity and projections.