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Vector Algebra question

2024 · 9 Apr · Shift 2 · Q42
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  5. /2024 · 9 Apr · Shift 2 · Q42

Vector Algebra question

2024 · 9 Apr · Shift 2 · Q42

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Between the following two statements: Statement I : Let a⃗=i^+2j^−3k^\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}a=i^+2j^​−3k^ and b⃗=2i^+j^−k^\vec{b}=2 \hat{i}+\hat{j}-\hat{k}b=2i^+j^​−k^. Then the vector r⃗\vec{r}r satisfying a⃗×r⃗=a⃗×b⃗\vec{a} \times \vec{r}=\vec{a} \times \vec{b}a×r=a×b and a⃗⋅r⃗=0\vec{a} \cdot \vec{r}=0a⋅r=0 is of magnitude 10\sqrt{10}10​. Statement II : In a triangle ABC,cos⁡2A+cos⁡2B+cos⁡2C≥−32A B C, \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2}ABC,cos2A+cos2B+cos2C≥−23​.
  1. A
    Both Statement I and Statement II are correct.
  2. B
    Both Statement I and Statement II are incorrect.
  3. C
    Statement I is correct but Statement II is incorrect.
  4. D
    Statement I is incorrect but Statement II is correct.
View written solutionFree

Correct answer: D

  1. Statement I

We are given a⃗=i^+2j^−3k^,b⃗=2i^+j^−k^\vec a=\hat i+2\hat j-3\hat k,\qquad \vec b=2\hat i+\hat j-\hat ka=i^+2j^​−3k^,b=2i^+j^​−k^ and need the vector r⃗\vec rr satisfying a⃗×r⃗=a⃗×b⃗\vec a\times \vec r=\vec a\times \vec ba×r=a×b and a⃗⋅r⃗=0.\vec a\cdot \vec r=0.a⋅r=0.

Step 1: Use the cross product condition

From a⃗×r⃗=a⃗×b⃗,\vec a\times \vec r=\vec a\times \vec b,a×r=a×b, we get a⃗×(r⃗−b⃗)=0.\vec a\times (\vec r-\vec b)=0.a×(r−b)=0. So r⃗−b⃗\vec r-\vec br−b must be parallel to a⃗\vec aa. Hence r⃗=b⃗+λa⃗\vec r=\vec b+\lambda \vec ar=b+λa for some scalar λ\lambdaλ.

Step 2: Use the dot product condition

Now a⃗⋅r⃗=0\vec a\cdot \vec r=0a⋅r=0 becomes a⃗⋅(b⃗+λa⃗)=0\vec a\cdot (\vec b+\lambda \vec a)=0a⋅(b+λa)=0 a⃗⋅b⃗+λ(a⃗⋅a⃗)=0.\vec a\cdot \vec b+\lambda (\vec a\cdot \vec a)=0.a⋅b+λ(a⋅a)=0.

Compute: a⃗⋅b⃗=(1)(2)+(2)(1)+(−3)(−1)=2+2+3=7,\vec a\cdot \vec b=(1)(2)+(2)(1)+(-3)(-1)=2+2+3=7,a⋅b=(1)(2)+(2)(1)+(−3)(−1)=2+2+3=7, a⃗⋅a⃗=12+22+(−3)2=1+4+9=14.\vec a\cdot \vec a=1^2+2^2+(-3)^2=1+4+9=14.a⋅a=12+22+(−3)2=1+4+9=14. So 7+14λ=0  ⟹  λ=−12.7+14\lambda=0\implies \lambda=-\frac12.7+14λ=0⟹λ=−21​.

Thus r⃗=b⃗−12a⃗.\vec r=\vec b-\frac12\vec a.r=b−21​a.

Step 3: Find r⃗\vec rr

=\left(2-\frac12,1-1,-1+\frac32\right) =\left(\frac32,0,\frac12\right).$$ ### Step 4: Magnitude of $\vec r$ $$|\vec r|=\sqrt{\left(\frac32\right)^2+0^2+\left(\frac12\right)^2} =\sqrt{\frac94+\frac14} =\sqrt{\frac{10}{4}} =\frac{\sqrt{10}}{2}.$$ But the statement says magnitude is $\sqrt{10}$, which is false. So, **Statement I is incorrect**. --- 2. **Statement II** We need to check whether in a triangle $ABC$, $$\cos 2A+\cos 2B+\cos 2C\ge -\frac32.$$ Since $$A+B+C=\pi,$$ there is a standard identity: $$\cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C.$$ So we need the minimum possible value of $$-1-4\cos A\cos B\cos C.$$ In a triangle, the known bound is $$\cos A\cos B\cos C\le \frac18,$$ with equality for $A=B=C=\frac\pi3$. Also, for an obtuse triangle this product can be negative, making the expression larger, not smaller. Thus the minimum occurs when the product is as large as possible, i.e. $\frac18$. Therefore, $$\cos 2A+\cos 2B+\cos 2C\ge -1-4\left(\frac18\right) =-1-\frac12 =-\frac32.$$ Equality holds for an equilateral triangle. So, **Statement II is correct**. --- 3. **Conclusion** - Statement I: Incorrect - Statement II: Correct Hence the correct option is $$\boxed{\text{D}}$$
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