To solve for the expression 2a2+3b2+4c2, follow these steps: Given that
^
e
=a
^
i
+b
^
j
+c
^
k
is a unit vector, we have: a2+b2+c2=1 The vector
^
e
is coplanar with the vectors
^
i
−3
^
j
+5
^
k
and 3
^
i
+
^
j
−5
^
k
. The condition for coplanarity implies the determinant formed by these vectors should be zero: |
a
b
c
1
−3
5
3
1
−5
|=0 Calculating the determinant gives: a(10)−b(−20)+c(10)=0‌‌⇒‌‌10a+20b+10c=0 Additionally, since
^
e
is perpendicular to
^
i
+
^
j
+
^
k
, we have: a+b+c=0 Solving the linear equations, we use the relation derived from coplanarity: 10a+20b+10c=0‌‌⇒‌‌a+2b+c=0 Together with a+b+c=0, it follows: ‌
a
1
=‌
−b
0
=‌
c
−1
=λ Thus:
a=λ,‌‌b=0,‌‌c=−λ Substituting into the unit vector condition: a2+b2+c2=1‌‌⇒‌‌λ2+0+(−λ)2=1‌‌⇒‌‌2λ2=1 Hence: λ2=‌