If a vector - x- 3k- is doubled in magnitude- then it becomes 4- -4x -2- - 2k- The values of x is are

Since, the vector î+ x ĵ + 3k̂ is doubled and its magnitude become equal to vector 4î +(4x 2)ĵ+ 2k̂ Therefore, 2|î +x ĵ + 3k̂| = |4 î + (4x 2)ĵ + 2k̂| 2 √(1 ...

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Standard XIII

Physics

Introduction

If a vector î...

Question

If a vector $^i + x^j + 3^k$ is doubled in magnitude, then it becomes $4^i + (4 x - 2)^j + 2^k$ . The value(s) of x is (are)

- \frac{2}{3}

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\frac{1}{3}

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\frac{2}{3}

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Solution

The correct option is D 2
Since, the vector $^i + x^j + 3^k$ is doubled and its magnitude become equal to vector $4^i + (4 x - 2)^j + 2^k$
Therefore,
$2 |^i + x^j + 3^k | = | 4^i + (4 x - 2)^j + 2^k |$
$2 \sqrt{1 + x^{2} + 9} = \sqrt{16 + (4 x - 2)^{2} + 4}$
$40 + 4 x^{2} = 20 + (4 x - 2)^{2}$
$3 x^{2} - 4 x - 4 = 0$
$(x - 2) (3 x + 2) = 0$
$x = 2, - \frac{2}{3}$

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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)

The correct option is D 2Since, 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

If a vector $^i + x^j + 3^k$ is doubled in magnitude, then it becomes $4^i + (4 x - 2)^j + 2^k$ . The value(s) of x is (are)