To solve for the value of [tex]\( k \)[/tex] that makes the vector from point [tex]\( A (1, -1, 3) \)[/tex] to point [tex]\( B (3, 0, 5) \)[/tex] perpendicular to the vector from point [tex]\( A \)[/tex] to point [tex]\( P (k, k, k) \)[/tex], we'll go through the following steps:
1. Determine the vector AB:
The vector from [tex]\( A \)[/tex] to [tex]\( B \)[/tex] is:
[tex]\[
\vec{AB} = B - A = (3 - 1, 0 - (-1), 5 - 3) = (2, 1, 2)
\][/tex]
2. Determine the vector AP:
The vector from [tex]\( A \)[/tex] to [tex]\( P \)[/tex] is:
[tex]\[
\vec{AP} = P - A = (k - 1, k + 1, k - 3)
\][/tex]
3. Set the dot product of [tex]\(\vec{AB}\)[/tex] and [tex]\(\vec{AP}\)[/tex] to zero:
Two vectors are perpendicular if and only if their dot product is zero. Therefore, we set up the equation:
[tex]\[
\vec{AB} \cdot \vec{AP} = 0
\][/tex]
Substituting the vectors [tex]\(\vec{AB} = (2, 1, 2)\)[/tex] and [tex]\(\vec{AP} = (k - 1, k + 1, k - 3)\)[/tex] into the dot product, we get:
[tex]\[
(2, 1, 2) \cdot (k - 1, k + 1, k - 3) = 0
\][/tex]
4. Calculate the dot product:
Compute the dot product term by term:
[tex]\[
2(k - 1) + 1(k + 1) + 2*(k - 3) = 0
\][/tex]
5. Simplify the expression:
Expand and simplify the terms:
[tex]\[
2k - 2 + k + 1 + 2k - 6 = 0
\][/tex]
Combine like terms:
[tex]\[
2k + k + 2k - 2 + 1 - 6 = 0
\][/tex]
[tex]\[
5k - 7 = 0
\][/tex]
6. Solve for [tex]\( k \)[/tex]:
Isolate [tex]\( k \)[/tex] by adding 7 to both sides:
[tex]\[
5k = 7
\][/tex]
Then divide by 5:
[tex]\[
k = \frac{7}{5}
\][/tex]
Hence, the value of [tex]\( k \)[/tex] that makes the vector from point [tex]\( A \)[/tex] to point [tex]\( B \)[/tex] perpendicular to the vector from point [tex]\( A \)[/tex] to point [tex]\( P \)[/tex] is [tex]\( \frac{7}{5} \)[/tex].
