We're given:
a=2i^−j^, b=−4j^+k^, c=i^+2k^
Since d is perpendicular to both a and b, it must be parallel to their cross product.
So, d=λ(a×b) for some scalar λ.
Writing the vectors with all three components:
a=2i^−1j^+0k^, b=0i^−4j^+1k^
a×b=i^20j^−1−4k^01
=i^[(−1)(1)−(0)(−4)]−j^[(2)(1)−(0)(0)]+k^[(2)(−4)−(−1)(0)]
=−i^−2j^−8k^
So, d=λ(−i^−2j^−8k^)
Using the condition c⋅d=34:
c⋅d=(i^+2k^)⋅λ(−i^−2j^−8k^)
=λ[(−1)+(0)+(−16)]
=−17λ
Setting this equal to 34:
−17λ=34
λ=−2
d=−2(−i^−2j^−8k^)=2i^+4j^+16k^
∣d∣=22+42+162
=4+16+256
=276
=269
∣d∣=269