Let a = i + j + k, b = -i - 8j + 2k and c = 4i + c2j + c3k be three vectors such that b × a = c × a.

Q: Wait, why can't we just say $\vec{b} = \vec{c}$? If $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$, doesn't that mean $\vec{b}$ and $\vec{c}$ are the same?

That is a classic trap! The cross product is not injective. If $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$, it only implies that the difference $(\vec{b} - \vec{c})$ is parallel to $\vec{a}$. Think of it geometrically: the cross product only captures the component of the vector perpendicular to $\vec{a}$. The component parallel to $\vec{a}$ is 'lost' in the cross product, which is why $\vec{b}$ and $\vec{c}$ can differ by a multiple of $\vec{a}$.

Q: How do I know $\lambda$ is a scalar and not a vector? The equation $\vec{b} - \vec{c} = \lambda \vec{a}$ looks like it could be anything.

Great question! By definition, if two vectors are parallel, one must be a scalar multiple of the other. Since $\vec{a}$ is a fixed vector, $\lambda$ must be a scalar constant that scales the magnitude and potentially flips the direction of $\vec{a}$ to match the direction of $(\vec{b} - \vec{c})$. It cannot be a vector because that would change the directionality of the relationship.

Q: Is there a faster way to find $\tan^2 \theta$ without calculating $\cos^2 \theta$ first?

While you could potentially use the identity $\sin^2 \theta = 1 - \cos^2 \theta$ and then $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$, it is almost always safer and faster to calculate $\cos^2 \theta$ first and then use $\tan^2 \theta = \sec^2 \theta - 1$. It keeps the algebra clean and minimizes the chance of a sign error.

Question

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = -\hat{i} - 8\hat{j} + 2\hat{k}$ and $\vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k}$ be three vectors such that $\vec{b} 	imes \vec{a} = \vec{c} 	imes \vec{a}$. If the angle between the vector $\vec{c}$ and the vector $3\hat{i} + 4\hat{j} + \hat{k}$ is $	heta$, then the greatest integer less than or equal to $	an^2 	heta$ is :

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given three vectors:
$\vec{a} = \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = -\hat{i} - 8\hat{j} + 2\hat{k}$
$\vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k}$

The governing condition is $\vec{b} 	imes \vec{a} = \vec{c} 	imes \vec{a}$. Instead of computing determinants, we rearrange the equation:
$$(\vec{b} - \vec{c}) 	imes \vec{a} = \vec{0}$$

This implies that the vector $(\vec{b} - \vec{c})$ is parallel to $\vec{a}$. Therefore, we can express this relationship using a scalar $\lambda$:
$$\vec{b} - \vec{c} = \lambda \vec{a} \quad \Rightarrow \quad \vec{c} = \vec{b} - \lambda \vec{a}$$

### Determining the Unknown Vector

We utilize the $x$-components of the vectors to solve for $\lambda$. Given $c_x = 4$, $b_x = -1$, and $a_x = 1$:
$$4 = -1 - \lambda(1)$$
$$\lambda = -5$$

Substituting $\lambda = -5$ back into the expression for $\vec{c}$:
$$\vec{c} = \vec{b} + 5\vec{a}$$
$$\vec{c} = (-\hat{i} - 8\hat{j} + 2\hat{k}) + 5(\hat{i} + \hat{j} + \hat{k})$$
$$\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$$

### Calculating the Angle

We now find the angle $	heta$ between $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$ and $\vec{d} = 3\hat{i} + 4\hat{j} + \hat{k}$. Using the dot product formula $\cos 	heta = \frac{\vec{c} \cdot \vec{d}}{|\vec{c}| |\vec{d}|}$:

$$\vec{c} \cdot \vec{d} = (4)(3) + (-3)(4) + (7)(1) = 12 - 12 + 7 = 7$$

The magnitudes are:
$$|\vec{c}| = \sqrt{4^2 + (-3)^2 + 7^2} = \sqrt{16 + 9 + 49} = \sqrt{74}$$
$$|\vec{d}| = \sqrt{3^2 + 4^2 + 1^2} = \sqrt{9 + 16 + 1} = \sqrt{26}$$

### Final Calculation

Substituting these values into the cosine expression:
$$\cos 	heta = \frac{7}{\sqrt{74} \sqrt{26}} \quad \Rightarrow \quad \cos^2 	heta = \frac{49}{74 	imes 26} = \frac{49}{1924}$$

Using the identity $	an^2 	heta = \frac{1}{\cos^2 	heta} - 1$:
$$	an^2 	heta = \frac{1924}{49} - 1 = \frac{1875}{49}$$

Calculating the numerical value:
$$	an^2 	heta \approx 38.265$$

The greatest integer less than or equal to this value is **38**.

Enter Numerical Value:

Visualized Solution

Identify Given Vectors $a$ and $b$

Analyze the Cross Product Condition

Using the Collinearity Property

Solving for $λ$

Finding Components $c_{2}$ and $c_{3}$

Define Vector $d$ and Angle $θ$

Calculate Dot Product $c \cdot d$

Calculate Magnitudes $∣ c ∣$ and $∣ d ∣$

Find $cos^{2} θ$

Calculate $tan^{2} θ$

Final Answer: Greatest Integer Value

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Determining the Unknown Vector

Calculating the Angle

Final Calculation

Similar Questions

Let $a = 3 \hat{i} + 2 \hat{j} + \hat{k}$ , $b = 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $c$ be a vector such that $(a + b) \times c = 2 (a \times b) + 24 \hat{j} - 6 \hat{k}$ and $(a - b + \hat{i}) \cdot c = - 3$ . Then $∣ c ∣^{2}$ is equal to _______.

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . If $c$ is a vector such that $a \cdot c = ∣ c ∣, ∣ c - a ∣ = 22$ and the angle between $(a \times b)$ and $c$ is $3 0^{\circ}$ , then $∣ (a \times b) \times c ∣ =$

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Let $a, b, c$ be three vectors such that $∣ a ∣ = 31, 4∣ b ∣ = ∣ c ∣ = 2$ and $2 (a \times b) = 3 (c \times a)$ . If the angle between $b$ and $c$ is $2 π /3$ , then $(\frac{a \times c}{a \cdot b})^{2}$ is equal to

Let $a = \hat{i} + 5 \hat{j} + α \hat{k}$ , $b = \hat{i} + 3 \hat{j} + β \hat{k}$ and $c = - \hat{i} + 2 \hat{j} - 3 \hat{k}$ be three vectors such that, $∣ b \times c ∣ = 53$ and $a$ is perpendicular to $b$ . Then the greatest amongst the values of $∣ a ∣^{2}$ is .

Let $a = - \hat{i} + 2 \hat{j} + 2 \hat{k}, b = 8 \hat{i} + 7 \hat{j} - 3 \hat{k}$ and $c$ be vector such that $a \times c = b$ . If $c \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$ , then $∣ a + c ∣^{2}$ is equal to :

Let $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ be two vectors such that $∣ a ∣ = 1$ ; $a \cdot b = 2$ and $∣ b ∣ = 4$ . If $c = 2 (a \times b) - 3 b$ , then the angle between $b$ and $c$ is equal to :

Let $a$ and $b$ be two vectors such that $∣ a ∣ = 1, ∣ b ∣ = 4$ and $a \cdot b = 2$ . If $c = (2 a \times b) - 3 b$ and the angle between $b$ and $c$ is $α$ , then $192 sin^{2} α$ is equal to _______.

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}, b = \hat{i} + \hat{j}$ and $c = a \times b$ . Let $d$ be a vector such that $∣ d - a ∣ = 11, ∣ c \times d ∣ = 3$ and the angle between $c$ and $d$ is $\frac{π}{4}$ . Then $a \cdot d$ is equal to

Enter Numerical Value:

Visualized Solution

Identify Given Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Analyze the Cross Product Condition

Using the Collinearity Property

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Finding Components .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c2​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c3​

Define Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d and Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Calculate Dot Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c⋅d

Calculate Magnitudes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣c∣ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣d∣

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos2θ

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan2θ