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Vector Algebra question

2022 · 24 Jun · Shift 1 · Q33
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  5. /2022 · 24 Jun · Shift 1 · Q33

Vector Algebra question

2022 · 24 Jun · Shift 1 · Q33

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a^\widehat aa, b^\widehat bb be unit vectors. If c→\overrightarrow cc be a vector such that the angle between a^\widehat aa and c→\overrightarrow cc is π12{\pi \over {12}}12π​, and b^=c→+2(c→×a^)\widehat b = \overrightarrow c + 2\left( {\overrightarrow c \times \widehat a} \right)b=c+2(c×a), then ∣6c→∣2{\left| {6\overrightarrow c } \right|^2}​6c​2 is equal to :
  1. A
    6(3−3)6\left( {3 - \sqrt 3 } \right)6(3−3​)
  2. B
    3+33 + \sqrt 33+3​
  3. C
    6(3+3)6\left( {3 + \sqrt 3 } \right)6(3+3​)
  4. D
    6(3+1)6\left( {\sqrt 3 + 1} \right)6(3​+1)
View written solutionFree

Correct answer: C

  1. Given data
  • a^,b^\hat a, \hat ba^,b^ are unit vectors.
  • Angle between a^\hat aa^ and c⃗\vec cc is π12=15∘\dfrac{\pi}{12}=15^\circ12π​=15∘.
  • Relation: b^=c⃗+2(c⃗×a^)\hat b=\vec c+2(\vec c\times \hat a)b^=c+2(c×a^)
  • Since b^\hat bb^ is a unit vector, ∣b^∣=1|\hat b|=1∣b^∣=1

We need to find ∣6c⃗∣2=36∣c⃗∣2.|6\vec c|^2=36|\vec c|^2.∣6c∣2=36∣c∣2.

  1. Use the magnitude of b^\hat bb^

Let ∣c⃗∣=m.|\vec c|=m.∣c∣=m.

Then b^=c⃗+2(c⃗×a^).\hat b=\vec c+2(\vec c\times \hat a).b^=c+2(c×a^).

Now note that c⃗\vec cc is perpendicular to (c⃗×a^)(\vec c\times \hat a)(c×a^), because a cross product is perpendicular to each of the vectors involved. Hence, c⃗⋅(c⃗×a^)=0.\vec c\cdot (\vec c\times \hat a)=0.c⋅(c×a^)=0.

Therefore, ∣b^∣2=∣c⃗∣2+4∣c⃗×a^∣2.|\hat b|^2=|\vec c|^2+4|\vec c\times \hat a|^2.∣b^∣2=∣c∣2+4∣c×a^∣2.

Since ∣b^∣=1|\hat b|=1∣b^∣=1, 1=m2+4∣c⃗×a^∣2.1=m^2+4|\vec c\times \hat a|^2.1=m2+4∣c×a^∣2.

  1. Compute ∣c⃗×a^∣|\vec c\times \hat a|∣c×a^∣

We know ∣c⃗×a^∣=∣c⃗∣ ∣a^∣sin⁡15∘=msin⁡15∘,|\vec c\times \hat a|=|\vec c|\,|\hat a|\sin 15^\circ = m\sin 15^\circ,∣c×a^∣=∣c∣∣a^∣sin15∘=msin15∘, because ∣a^∣=1|\hat a|=1∣a^∣=1.

So, 1=m2+4m2sin⁡215∘.1=m^2+4m^2\sin^2 15^\circ.1=m2+4m2sin215∘.

Thus, 1=m2(1+4sin⁡215∘).1=m^2(1+4\sin^2 15^\circ).1=m2(1+4sin215∘).

Hence, m2=11+4sin⁡215∘.m^2=\frac{1}{1+4\sin^2 15^\circ}.m2=1+4sin215∘1​.

  1. Evaluate sin⁡215∘\sin^2 15^\circsin215∘

Use sin⁡15∘=sin⁡(45∘−30∘)=6−24.\sin 15^\circ=\sin(45^\circ-30^\circ)=\frac{\sqrt6-\sqrt2}{4}.sin15∘=sin(45∘−30∘)=46​−2​​.

So,

=\frac{6+2-2\sqrt{12}}{16} =\frac{8-4\sqrt3}{16} =\frac{2-\sqrt3}{4}.$$ Therefore, $$4\sin^2 15^\circ=2-\sqrt3.$$ So, $$1+4\sin^2 15^\circ=1+(2-\sqrt3)=3-\sqrt3.$$ Thus, $$m^2=\frac{1}{3-\sqrt3}.$$ Rationalizing, $$m^2=\frac{3+\sqrt3}{6}.$$ --- 5. **Find $|6\vec c|^2$** $$|6\vec c|^2=36|\vec c|^2=36m^2=36\cdot \frac{3+\sqrt3}{6}=6(3+\sqrt3).$$ --- 6. **Compare with options** The value is $$6(3+\sqrt3),$$ which is **Option C**. --- 7. **Comparison with stored correct answer** Stored correct answer: **C** Our derived answer: **C** So the derived answer agrees with the stored correct answer.
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