To find the value of [tex]\(\lambda\)[/tex] given that the perpendicular distance of the point [tex]\((6, 5, 8)\)[/tex] from the y-axis is 5 units, we need to understand the concept of the distance from a point to an axis.
The y-axis in a 3D coordinate system is the set of all points where the x and z coordinates are both zero, i.e., [tex]\((0, y, 0)\)[/tex]. The perpendicular distance from any point [tex]\((x, y, z)\)[/tex] to the y-axis is determined by the horizontal distance in the xz-plane, ignoring the y-coordinate.
This distance is given by:
[tex]\[
\text{Distance} = \sqrt{x^2 + z^2}
\][/tex]
Here is how we can break this down step-by-step:
1. Identify the coordinates:
The given point is [tex]\((6, 5, 8)\)[/tex]. We will use [tex]\(x = 6\)[/tex] and [tex]\(z = 8\)[/tex]. The y-coordinate (5) does not affect the distance to the y-axis.
2. Formulate the distance calculation:
The formula for the perpendicular distance from a point [tex]\((x, y, z)\)[/tex] to the y-axis is:
[tex]\[
\text{Distance} = \sqrt{x^2 + z^2}
\][/tex]
3. Plug in the values:
Substitute [tex]\(x = 6\)[/tex] and [tex]\(z = 8\)[/tex] into the formula:
[tex]\[
\text{Distance} = \sqrt{6^2 + 8^2}
\][/tex]
4. Calculate the squares:
[tex]\[
6^2 = 36 \quad \text{and} \quad 8^2 = 64
\][/tex]
5. Add the squares:
[tex]\[
36 + 64 = 100
\][/tex]
6. Take the square root:
[tex]\[
\sqrt{100} = 10
\][/tex]
Thus, the perpendicular distance from the point [tex]\((6, 5, 8)\)[/tex] to the y-axis is indeed 10 units, not 5 units as given in the problem statement. Hence, the value of [tex]\(\lambda\)[/tex] is:
[tex]\[
\lambda = 10
\][/tex]
Therefore, the value of [tex]\(\lambda\)[/tex] is [tex]\(10\)[/tex].
