Question

Let a₁,a₂,a₃ ∈ R-{0} and a₁ + a₂cos2x + a₃sin²x = 0 ∀ x ∈ R, then-

• A. vectors $\overrightarrow{a}$ = a₁$\hat{i}$ + a₂$\hat{j}$ + a₃$\hat{k}$ and $\overrightarrow{b}$ = 4$\hat{i}$ + 2$\hat{j}$ + $\hat{k}$ are perpendicular
• B. vectors $\overrightarrow{a}$ = a₁$\hat{i}$ + a₂$\hat{j}$ + a₃$\hat{k}$ and $\overrightarrow{b}$ = $\hat{i}$ + $\hat{j}$ + 2$\hat{k}$ are parallel to each other
• C. if vector $\overrightarrow{a}$ = a₁$\hat{i}$ + a₂$\hat{j}$ + a₃$\hat{k}$ is of length $\sqrt{6}$, then one of ordered triplet (a₁,a₂,a₃) = (1,-1,-2)
• D. if 2a₁ + 3a₂ + 6a₃ = 26, then |a₁$\hat{i}$ + a₂$\hat{j}$ + a₃$\hat{k}$| = 2$\sqrt{6}$

Answer 1

Answer: (A), (C), (D)

A, C, D

Answer 2

The given equation is $a_1 + a_2 \cos(2x) + a_3 \sin^2(x) = 0$ for all $x \in R$, where $a_1, a_2, a_3 \in R - \{0\}$.

We use the trigonometric identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$. Substituting this into the equation, we get:

$a_1 + a_2 \cos(2x) + a_3 \left(\frac{1 - \cos(2x)}{2}\right) = 0$

$a_1 + a_2 \cos(2x) + \frac{a_3}{2} - \frac{a_3}{2} \cos(2x) = 0$

Rearranging the terms, we group the constant terms and the terms involving $\cos(2x)$:

$\left(a_1 + \frac{a_3}{2}\right) + \left(a_2 - \frac{a_3}{2}\right) \cos(2x) = 0$

This equation must hold for all $x \in R$. Since $\cos(2x)$ is not a constant function, for this linear combination of a constant and $\cos(2x)$ to be identically zero, the coefficients of the constant term and $\cos(2x)$ must both be zero.

So, we have two equations:

1. $a_1 + \frac{a_3}{2} = 0 \implies 2a_1 + a_3 = 0 \implies a_3 = -2a_1$

2. $a_2 - \frac{a_3}{2} = 0 \implies 2a_2 - a_3 = 0 \implies a_3 = 2a_2$

From these two equations, we have $a_3 = -2a_1$ and $a_3 = 2a_2$. Equating the expressions for $a_3$, we get $-2a_1 = 2a_2$, which simplifies to $a_1 = -a_2$, or $a_2 = -a_1$. Then $a_3 = 2a_2 = 2(-a_1) = -2a_1$.

So the relationship between $a_1, a_2, a_3$ is $a_2 = -a_1$ and $a_3 = -2a_1$. Since $a_1 \in R - \{0\}$, we can express $a_2$ and $a_3$ in terms of $a_1$. Let $a_1 = \lambda$, where $\lambda \in R - \{0\}$. Then $a_2 = -\lambda$ and $a_3 = -2\lambda$. The ordered triplet $(a_1, a_2, a_3)$ is $(\lambda, -\lambda, -2\lambda) = \lambda(1, -1, -2)$. This means $(a_1, a_2, a_3)$ is proportional to $(1, -1, -2)$. The vector $\overrightarrow{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} = \lambda\hat{i} - \lambda\hat{j} - 2\lambda\hat{k} = \lambda(\hat{i} - \hat{j} - 2\hat{k})$.

Now let's check the given options:

(A) vectors $\overrightarrow{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\overrightarrow{b} = 4\hat{i} + 2\hat{j} + \hat{k}$ are perpendicular. The dot product $\overrightarrow{a} \cdot \overrightarrow{b} = (a_1)(4) + (a_2)(2) + (a_3)(1) = 4a_1 + 2a_2 + a_3$. Substitute $a_2 = -a_1$ and $a_3 = -2a_1$: $4a_1 + 2(-a_1) + (-2a_1) = 4a_1 - 2a_1 - 2a_1 = 0$. Since the dot product is zero, the vectors are perpendicular. Option (A) is correct.

(B) vectors $\overrightarrow{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\overrightarrow{b} = \hat{i} + \hat{j} + 2\hat{k}$ are parallel to each other. $\overrightarrow{a} = \lambda(\hat{i} - \hat{j} - 2\hat{k})$. For $\overrightarrow{a}$ and $\overrightarrow{b}$ to be parallel, $\overrightarrow{a}$ must be a scalar multiple of $\overrightarrow{b}$. $\lambda(\hat{i} - \hat{j} - 2\hat{k}) = c(\hat{i} + \hat{j} + 2\hat{k})$ for some scalar $c$. Comparing components: $\lambda = c$, $-\lambda = c$, $-2\lambda = 2c$. From $\lambda = c$ and $-\lambda = c$, we get $\lambda = -\lambda$, so $2\lambda = 0$, which implies $\lambda = 0$. However, $a_1 = \lambda \neq 0$. Thus, $\overrightarrow{a}$ and $\overrightarrow{b}$ are not parallel. Option (B) is incorrect.

(C) if vector $\overrightarrow{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ is of length $\sqrt{6}$, then one of ordered triplet $(a_1, a_2, a_3) = (1, -1, -2)$. The length of $\overrightarrow{a}$ is $|\overrightarrow{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$. Substitute $a_1 = \lambda$, $a_2 = -\lambda$, $a_3 = -2\lambda$: $|\overrightarrow{a}| = \sqrt{\lambda^2 + (-\lambda)^2 + (-2\lambda)^2} = \sqrt{\lambda^2 + \lambda^2 + 4\lambda^2} = \sqrt{6\lambda^2} = |\lambda|\sqrt{6}$. Given $|\overrightarrow{a}| = \sqrt{6}$, we have $|\lambda|\sqrt{6} = \sqrt{6}$, which implies $|\lambda| = 1$. So $\lambda = 1$ or $\lambda = -1$. If $\lambda = 1$, $(a_1, a_2, a_3) = (1, -1, -2)$. If $\lambda = -1$, $(a_1, a_2, a_3) = (-1, 1, 2)$. The option states that one of the ordered triplet is $(1, -1, -2)$. This is true when $\lambda = 1$. Option (C) is correct.

(D) if $2a_1 + 3a_2 + 6a_3 = 26$, then $|a_1\hat{i} + a_2\hat{j} + a_3\hat{k}| = 2\sqrt{6}$. Substitute $a_1 = \lambda$, $a_2 = -\lambda$, $a_3 = -2\lambda$ into the equation $2a_1 + 3a_2 + 6a_3 = 26$: $2(\lambda) + 3(-\lambda) + 6(-2\lambda) = 26$. $2\lambda - 3\lambda - 12\lambda = 26$. $-13\lambda = 26$. $\lambda = \frac{26}{-13} = -2$. Now calculate the magnitude of $\overrightarrow{a}$: $|\overrightarrow{a}| = |\lambda|\sqrt{6}$. With $\lambda = -2$, $|\lambda| = |-2| = 2$. $|\overrightarrow{a}| = 2\sqrt{6}$. The option states that $|a_1\hat{i} + a_2\hat{j} + a_3\hat{k}| = 2\sqrt{6}$. This is true. Option (D) is correct.

All options (A), (C), and (D) are correct.