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Vector Algebra question

2022 · Shift 2 · Q31
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Vector Algebra question

2022 · Shift 2 · Q31

JEE AdvancedMathematicsVector AlgebraMultiple correct+4 / −2
Let ı^,ȷ^\hat{\imath}, \hat{\jmath}^,^​ and k^\hat{k}k^ be the unit vectors along the three positive coordinate axes. Let a⃗=3ı^+ȷ^−k^, b⃗=ı^+b2ȷ^+b3k^,b2,b3∈R, c⃗=c1ı^+c2ȷ^+c3k^,c1,c2,c3∈R\begin{aligned} & \vec{a}=3 \hat{\imath}+\hat{\jmath}-\hat{k} \text {, } \\ & \vec{b}=\hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}, \quad b_{2}, b_{3} \in \mathbb{R} \text {, } \\ & \vec{c}=c_{1} \hat{\imath}+c_{2} \hat{\jmath}+c_{3} \hat{k}, \quad c_{1}, c_{2}, c_{3} \in \mathbb{R} \end{aligned}​a=3^+^​−k^, b=^+b2​^​+b3​k^,b2​,b3​∈R, c=c1​^+c2​^​+c3​k^,c1​,c2​,c3​∈R​ be three vectors such that b2b3>0,a⃗⋅b⃗=0b_{2} b_{3}\gt 0, \vec{a} \cdot \vec{b}=0b2​b3​>0,a⋅b=0 and (0−c3c2c30−c1−c2c10)(1b2b3)=(3−c11−c2−1−c3).\left(\begin{array}{ccc} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{array}\right)\left(\begin{array}{l} 1 \\ b_{2} \\ b_{3} \end{array}\right)=\left(\begin{array}{r} 3-c_{1} \\ 1-c_{2} \\ -1-c_{3} \end{array}\right) .​0c3​−c2​​−c3​0c1​​c2​−c1​0​​​1b2​b3​​​=​3−c1​1−c2​−1−c3​​​. Then, which of the following is/are TRUE?
  1. A
    a⃗⋅c⃗=0\vec{a} \cdot \vec{c}=0a⋅c=0
  2. B
    b⃗⋅c⃗=0\vec{b} \cdot \vec{c}=0b⋅c=0
  3. C
    ∣b⃗∣>10|\vec{b}|\gt \sqrt{10}∣b∣>10​
  4. D
    ∣c⃗∣≤11|\vec{c}| \leq \sqrt{11}∣c∣≤11​
View written solutionFree

Correct answer: B, C, D

  1. Write the vectors explicitly

a⃗=(3,1,−1),b⃗=(1,b2,b3),c⃗=(c1,c2,c3).\vec a=(3,1,-1),\qquad \vec b=(1,b_2,b_3),\qquad \vec c=(c_1,c_2,c_3).a=(3,1,−1),b=(1,b2​,b3​),c=(c1​,c2​,c3​).

Given:

  • b2b3>0b_2b_3>0b2​b3​>0
  • a⃗⋅b⃗=0\vec a\cdot \vec b=0a⋅b=0
0 & -c_3 & c_2\\ c_3 & 0 & -c_1\\ -c_2 & c_1 & 0 \end{pmatrix} \begin{pmatrix}1\\ b_2\\ b_3\end{pmatrix} = \begin{pmatrix}3-c_1\\ 1-c_2\\ -1-c_3\end{pmatrix}.$$ --- 2. **Interpret the matrix equation** The skew-symmetric matrix $$\begin{pmatrix} 0 & -c_3 & c_2\\ c_3 & 0 & -c_1\\ -c_2 & c_1 & 0 \end{pmatrix}$$ represents the cross product with $\vec c$, i.e. $$[\vec c]_\times \vec b=\vec c\times \vec b.$$ So the given equation is $$\vec c\times \vec b=(3-c_1,1-c_2,-1-c_3).$$ But $$(3-c_1,1-c_2,-1-c_3)=\vec a-\vec c.$$ Thus, $$\boxed{\vec c\times \vec b=\vec a-\vec c.}$$ Hence, $$\boxed{\vec a=\vec c+\vec c\times \vec b.}$$ --- 3. **Use dot product with $\vec b$** Take dot product of $$\vec a=\vec c+\vec c\times \vec b$$ with $\vec b$: $$\vec a\cdot \vec b=\vec c\cdot \vec b+(\vec c\times \vec b)\cdot \vec b.$$ Now $(\vec c\times \vec b)\cdot \vec b=0$, so $$0=\vec a\cdot \vec b=\vec c\cdot \vec b.$$ Therefore, $$\boxed{\vec b\cdot \vec c=0.}$$ So **Option B is true**. --- 4. **Use the condition $\vec a\cdot \vec b=0$ to find relation between $b_2,b_3$** $$\vec a\cdot \vec b=(3,1,-1)\cdot(1,b_2,b_3)=3+b_2-b_3=0.$$ So $$b_3=b_2+3.$$ Since $b_2b_3>0$, we need $$b_2(b_2+3)>0.$$ This gives $$b_2>0 \quad \text{or} \quad b_2<-3.$$ Now $$|\vec b|^2=1+b_2^2+b_3^2=1+b_2^2+(b_2+3)^2 =2b_2^2+6b_2+10.

Complete the square:

But allowed values are b2>0b_2>0b2​>0 or b2<−3b_2<-3b2​<−3, so the minimum is approached near the boundary values 000 and −3-3−3 (not included): ∣b⃗∣2>10.|\vec b|^2>10.∣b∣2>10. Hence ∣b⃗∣>10.\boxed{|\vec b|>\sqrt{10}.}∣b∣>10​.​

So Option C is true.

  1. Find ∣c⃗∣|\vec c|∣c∣ using orthogonality

From a⃗=c⃗+c⃗×b⃗,\vec a=\vec c+\vec c\times \vec b,a=c+c×b, and since c⃗⋅(c⃗×b⃗)=0,\vec c\cdot(\vec c\times \vec b)=0,c⋅(c×b)=0, we get by squaring magnitudes: ∣a⃗∣2=∣c⃗∣2+∣c⃗×b⃗∣2.|\vec a|^2=|\vec c|^2+|\vec c\times \vec b|^2.∣a∣2=∣c∣2+∣c×b∣2. Thus, ∣c⃗∣2≤∣a⃗∣2.|\vec c|^2\le |\vec a|^2.∣c∣2≤∣a∣2. Now ∣a⃗∣2=32+12+(−1)2=11.|\vec a|^2=3^2+1^2+(-1)^2=11.∣a∣2=32+12+(−1)2=11. Therefore, ∣c⃗∣≤11.|\vec c|\le \sqrt{11}.∣c∣≤11​.

So Option D is true.

  1. Check Option A

Take dot product of a⃗=c⃗+c⃗×b⃗\vec a=\vec c+\vec c\times \vec ba=c+c×b with c⃗\vec cc:

Since (c⃗×b⃗)⋅c⃗=0(\vec c\times \vec b)\cdot \vec c=0(c×b)⋅c=0, we get a⃗⋅c⃗=∣c⃗∣2.\boxed{\vec a\cdot \vec c=|\vec c|^2.}a⋅c=∣c∣2.​ This is zero only if c⃗=0⃗\vec c=\vec 0c=0, which is not necessary.

So Option A is false.

  1. Final conclusion

The true statements are: B, C, D\boxed{\text{B, C, D}}B, C, D​

These match the stored correct answer.

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