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Vector Algebra question

2018 · 15 Apr · Shift 1 · Q38
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Vector Algebra question

2018 · 15 Apr · Shift 1 · Q38

JEE MainMathematicsVector AlgebraMCQ+4 / −1
If a→,  b→,\overrightarrow a ,\,\,\overrightarrow b ,a,b, and C→\overrightarrow CC are unit vectors such that a→+2b→+2c→=0→,\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 ,a+2b+2c=0, then ∣a→×c→∣\left| {\overrightarrow a \times \overrightarrow c } \right|​a×c​ is equal to :
  1. A
    154{{\sqrt {15} } \over 4}415​​
  2. B
    14{{1} \over {4}}41​
  3. C
    1516{{15} \over {16}}1615​
  4. D
    1516{{\sqrt {15} } \over 16}1615​​
View written solutionFree

Correct answer: A

  1. Given condition

We have unit vectors a⃗,b⃗,c⃗\vec a, \vec b, \vec ca,b,c such that

a⃗+2b⃗+2c⃗=0⃗.\vec a + 2\vec b + 2\vec c = \vec 0.a+2b+2c=0.

So,

a⃗=−2(b⃗+c⃗).\vec a = -2(\vec b + \vec c).a=−2(b+c).
  1. Use magnitudes

Since a⃗\vec aa is a unit vector,

∣a⃗∣=1.|\vec a|=1.∣a∣=1.

Hence,

1=∣a⃗∣=2∣b⃗+c⃗∣⇒∣b⃗+c⃗∣=12.1 = |\vec a| = 2|\vec b + \vec c| \quad\Rightarrow\quad |\vec b + \vec c| = \frac12.1=∣a∣=2∣b+c∣⇒∣b+c∣=21​.

Now,

∣b⃗+c⃗∣2=∣b⃗∣2+∣c⃗∣2+2b⃗⋅c⃗.|\vec b + \vec c|^2 = |\vec b|^2 + |\vec c|^2 + 2\vec b\cdot \vec c.∣b+c∣2=∣b∣2+∣c∣2+2b⋅c.

Since b⃗\vec bb and c⃗\vec cc are unit vectors,

(12)2=1+1+2b⃗⋅c⃗.\left(\frac12\right)^2 = 1 + 1 + 2\vec b\cdot \vec c.(21​)2=1+1+2b⋅c.

So,

14=2+2b⃗⋅c⃗\frac14 = 2 + 2\vec b\cdot \vec c41​=2+2b⋅c 2b⃗⋅c⃗=−742\vec b\cdot \vec c = -\frac742b⋅c=−47​ b⃗⋅c⃗=−78.\vec b\cdot \vec c = -\frac78.b⋅c=−87​.
  1. Find a⃗⋅c⃗\vec a\cdot \vec ca⋅c

From

a⃗=−2(b⃗+c⃗),\vec a = -2(\vec b+\vec c),a=−2(b+c),

we get

a⃗⋅c⃗=−2(b⃗⋅c⃗+c⃗⋅c⃗).\vec a\cdot \vec c = -2(\vec b\cdot \vec c + \vec c\cdot \vec c).a⋅c=−2(b⋅c+c⋅c).

Since c⃗⋅c⃗=1\vec c\cdot \vec c = 1c⋅c=1,

a⃗⋅c⃗=−2(−78+1)=−2(18)=−14.\vec a\cdot \vec c = -2\left(-\frac78 + 1\right) = -2\left(\frac18\right) = -\frac14.a⋅c=−2(−87​+1)=−2(81​)=−41​.
  1. Use cross product formula

For unit vectors a⃗\vec aa and c⃗\vec cc,

∣a⃗×c⃗∣=1−(a⃗⋅c⃗)2.|\vec a \times \vec c| = \sqrt{1-(\vec a\cdot \vec c)^2}.∣a×c∣=1−(a⋅c)2​.

Thus,

∣a⃗×c⃗∣=1−(−14)2=1−116=1516=154.|\vec a \times \vec c| = \sqrt{1-\left(-\frac14\right)^2} = \sqrt{1-\frac1{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}.∣a×c∣=1−(−41​)2​=1−161​​=1615​​=415​​.
  1. Check options

The value is

154,\frac{\sqrt{15}}{4},415​​,

which matches Option A.

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