To determine the value of [tex]\( p \)[/tex] such that the vectors [tex]\( \vec{v_1} = p\vec{i} - 3\vec{j} \)[/tex] and [tex]\( \vec{v_2} = 3\vec{i} - 2\vec{j} \)[/tex] are perpendicular, we need to use the property that two vectors are perpendicular if and only if their dot product is zero.
Given:
[tex]\[
\vec{v_1} = p\vec{i} - 3\vec{j}
\][/tex]
[tex]\[
\vec{v_2} = 3\vec{i} - 2\vec{j}
\][/tex]
The dot product of two vectors [tex]\(\vec{a} = a_1\vec{i} + a_2\vec{j}\)[/tex] and [tex]\(\vec{b} = b_1\vec{i} + b_2\vec{j}\)[/tex] is given by:
[tex]\[
\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2
\][/tex]
Applying this to our vectors [tex]\(\vec{v_1}\)[/tex] and [tex]\(\vec{v_2}\)[/tex]:
[tex]\[
\vec{v_1} \cdot \vec{v_2} = (p\vec{i} - 3\vec{j}) \cdot (3\vec{i} - 2\vec{j}) = p \cdot 3 + (-3) \cdot (-2)
\][/tex]
Calculating the components:
[tex]\[
\vec{v_1} \cdot \vec{v_2} = 3p + 6
\][/tex]
For the vectors to be perpendicular, the dot product must be equal to zero:
[tex]\[
3p + 6 = 0
\][/tex]
Solving for [tex]\( p \)[/tex]:
[tex]\[
3p + 6 = 0
\][/tex]
[tex]\[
3p = -6
\][/tex]
[tex]\[
p = -2
\][/tex]
Thus, the value of [tex]\( p \)[/tex] that makes the vectors [tex]\( p\vec{i} - 3\vec{j} \)[/tex] and [tex]\( 3\vec{i} - 2\vec{j} \)[/tex] perpendicular is:
[tex]\[
\boxed{-2}
\][/tex]
