Vector Algebra

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=‌

1

2

‌‌⇒‌‌λ=±‌

1

√2

Therefore, the values for a,b, and c are:
a=‌

1

√2

,‌‌b=0,‌‌c=−‌

1

√2

Calculating 2a2+3b2+4c2 :
2(‌

1

2

)+3(0)+4(‌

1

2

)=1+2=3
Thus, 2a2+3b2+4c2=3.