a unit vactor a makes an angle of 45 with z axix and i+j+a is a unit vector , then a vector is?

To find the vector \( \mathbf{a} \) that makes a 45-degree angle with the z-axis and is part of the unit vector equation \( \mathbf{i} + \mathbf{j} + \mathbf{a} = \mathbf{u} \), where \( \mathbf{u} \) is also a unit vector, we can break this down step by step.

Understanding the Components

A unit vector has a magnitude of 1. The vector \( \mathbf{a} \) can be expressed in terms of its components along the x, y, and z axes as \( \mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k} \). Since \( \mathbf{a} \) makes a 45-degree angle with the z-axis, we can use trigonometric relationships to express \( a_z \) in terms of \( a_x \) and \( a_y \).

Using Trigonometry

The cosine of the angle between the vector and the z-axis can be expressed as:

• \( \cos(45^\circ) = \frac{a_z}{|\mathbf{a}|} \)

Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \), we have:

• \( a_z = \frac{|\mathbf{a}|}{\sqrt{2}} \)

Magnitude of the Vector

For a unit vector, the magnitude is given by:

• \( |\mathbf{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = 1 \)

Substituting \( a_z \) into the magnitude equation gives:

• \( 1 = \sqrt{a_x^2 + a_y^2 + \left(\frac{1}{\sqrt{2}}\right)^2} \)

Squaring both sides, we get:

• \( 1 = a_x^2 + a_y^2 + \frac{1}{2} \)

This simplifies to:

• \( a_x^2 + a_y^2 = \frac{1}{2} \)

Finding the Vector Components

Now, we know that \( \mathbf{i} + \mathbf{j} + \mathbf{a} = \mathbf{u} \) is a unit vector. Let's express \( \mathbf{u} \) in terms of its components:

• \( \mathbf{u} = (1 + a_x) \mathbf{i} + (1 + a_y) \mathbf{j} + a_z \mathbf{k} \)

The magnitude of \( \mathbf{u} \) must also equal 1:

• \( 1 = \sqrt{(1 + a_x)^2 + (1 + a_y)^2 + a_z^2} \)

Substituting \( a_z = \frac{1}{\sqrt{2}} \) into this equation leads to:

• \( 1 = \sqrt{(1 + a_x)^2 + (1 + a_y)^2 + \frac{1}{2}} \)

Solving the Equations

Squaring both sides again gives:

• \( 1 = (1 + a_x)^2 + (1 + a_y)^2 + \frac{1}{2} \)

Expanding this results in:

• \( 1 = 1 + 2a_x + a_x^2 + 1 + 2a_y + a_y^2 + \frac{1}{2} \)

Combining like terms leads to:

• \( 0 = 2a_x + a_x^2 + 2a_y + a_y^2 + \frac{1}{2} \)

Final Steps

At this point, we can substitute \( a_y = \sqrt{\frac{1}{2} - a_x^2} \) into the equation and solve for \( a_x \) and \( a_y \). This will yield the specific values for the components of vector \( \mathbf{a} \). The solutions will depend on the chosen values of \( a_x \) and \( a_y \) that satisfy the unit vector condition.

In conclusion, the vector \( \mathbf{a} \) can be expressed in terms of its components, and by solving the equations derived from the conditions given, we can find its specific values. This method illustrates how trigonometry and algebra work together to solve vector problems in three-dimensional space.