If a vector ^i+x^j+3^k is doubled in magnitude, then it becomes 4^i+(4x−2)^j+2^k. The value(s) of x is (are)
A
−23
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
13
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
23
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct options are A−23 D 2 Since, the vector ^i+x^j+3^k is doubled and its magnitude become equal to vector 4^i+(4x−2)^j+2^k Therefore, 2|^i+x^j+3^k|=|4^i+(4x−2)^j+2^k| 2√1+x2+9=√16+(4x−2)2+4 40+4x2=20+(4x−2)2 3x2−4x−4=0 (x−2)(3x+2)=0 x=2,−23