Let a = i + 2j + λk, b = 3i - 5j - λk, a · c = 7, 2b · c + 43 = 0, a × c = b × c. Then |a · b| is eq

Q: Wait, why can we just say $\vec{c} = \frac{1}{\mu}(\vec{a} - \vec{b})$? Is it possible that $\vec{c}$ is the zero vector?

That is a brilliant observation! In a formal proof, we must consider if $\vec{c} = \vec{0}$. However, if $\vec{c} = \vec{0}$, then $\vec{a} \cdot \vec{c}$ would be $0$, which contradicts the given condition $\vec{a} \cdot \vec{c} = 7$. Thus, $\vec{c}$ cannot be the zero vector, and our assumption of collinearity is mathematically sound.

Q: Why did we use $\mu$ instead of just a constant $k$?

Using $\mu$ is just a stylistic choice to represent the scalar constant of proportionality. You could use $k$, $\alpha$, or any symbol you like! The important part is recognizing that the cross product condition forces the vectors to be linearly dependent.

Q: I got stuck trying to find $\lambda$ individually. Do I really need to find $\lambda$?

Not at all! Notice that both the dot product $\vec{a} \cdot \vec{b}$ and the magnitude squares $|\vec{a}|^2$ and $|\vec{b}|^2$ depend only on $\lambda^2$. You only need to solve for $\lambda^2$ as a single block. Don't waste time calculating $\lambda = \pm 1$ unless the question specifically asks for it!

Q: Why does $(\vec{a} - \vec{b}) \times \vec{c} = \vec{0}$ imply they are parallel?

The cross product of two vectors $\vec{u}$ and $\vec{v}$ is defined as $|\vec{u}||\vec{v}| \sin(\theta) \hat{n}$. If the cross product is the zero vector, then the magnitude must be zero. Since we've established $\vec{c} \neq \vec{0}$ and $\vec{a} - \vec{b} \neq \vec{0}$, the only way for the magnitude to be zero is if $\sin(\theta) = 0$, which means $\theta = 0$ or $\pi$—the definition of parallel vectors!

Q: Is there a faster way to solve this without substituting $\vec{c}$?

You could try to solve for the components of $\vec{c} = (x, y, z)$ using the system of linear equations, but that would involve three variables plus $\lambda$. Substituting $\vec{c}$ in terms of $(\vec{a} - \vec{b})$ is the 'elegant' path because it reduces the complexity of the system significantly. Always look for ways to group vectors before expanding into components!

Question

Let $\vec{a} = \hat{i} + 2\hat{j} + \lambda \hat{k}$, $\vec{b} = 3\hat{i} - 5\hat{j} - \lambda \hat{k}$, $\vec{a} \cdot \vec{c} = 7$, $2\vec{b} \cdot \vec{c} + 43 = 0$, $\vec{a} 	imes \vec{c} = \vec{b} 	imes \vec{c}$. Then $|\vec{a} \cdot \vec{b}|$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two vectors, $\vec{a} = \hat{i} + 2\hat{j} + \lambda \hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} - \lambda \hat{k}$. We are also provided with a third vector $\vec{c}$ that satisfies specific dot product conditions.

Our objective is to determine the absolute value of the dot product, $|\vec{a} \cdot \vec{b}|$.

### The Power of Rearrangement

Many students see $\vec{a} 	imes \vec{c} = \vec{b} 	imes \vec{c}$ and immediately attempt to write out the determinant form. This often leads to unnecessary algebraic complexity.

Instead, we use the distributive property of the cross product. By moving all terms to one side, we obtain:

$$(\vec{a} - \vec{b}) 	imes \vec{c} = \vec{0}$$

This is the critical insight. In vector geometry, if the cross product of two non-zero vectors is the zero vector, the vectors must be collinear.

Consequently, we can express $\vec{c}$ as a scalar multiple of the difference between $\vec{a}$ and $\vec{b}$:

$$\vec{c} = \frac{1}{\mu}(\vec{a} - \vec{b})$$

### Bridging the Gap with Dot Products

We are given the conditions $\vec{a} \cdot \vec{c} = 7$ and $2\vec{b} \cdot \vec{c} + 43 = 0$. Substituting $\vec{c} = \frac{1}{\mu}(\vec{a} - \vec{b})$ into these equations yields:

$$|\vec{a}|^2 - \vec{a} \cdot \vec{b} = 7\mu$$

and

$$\vec{a} \cdot \vec{b} - |\vec{b}|^2 = -\frac{43\mu}{2}$$

Next, we calculate the magnitudes and the dot product in terms of $\lambda$:

$$|\vec{a}|^2 = 1^2 + 2^2 + \lambda^2 = 5 + \lambda^2$$

$$|\vec{b}|^2 = 3^2 + (-5)^2 + (-\lambda)^2 = 34 + \lambda^2$$

$$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-5) + (\lambda)(-\lambda) = 3 - 10 - \lambda^2 = -7 - \lambda^2$$

### The Final Convergence

Substituting these values into our system of equations, we obtain:

$$(5 + \lambda^2) - (-7 - \lambda^2) = 7\mu \Rightarrow 12 + 2\lambda^2 = 7\mu$$

$$(-7 - \lambda^2) - (34 + \lambda^2) = -\frac{43\mu}{2} \Rightarrow -41 - 2\lambda^2 = -\frac{43\mu}{2}$$

Multiplying the second equation by $-2$, we get $82 + 4\lambda^2 = 43\mu$. Solving the system $12 + 2\lambda^2 = 7\mu$ and $82 + 4\lambda^2 = 43\mu$ yields $\mu = 2$ and $\lambda^2 = 1$.

Finally, we calculate the target value:

$$|\vec{a} \cdot \vec{b}| = |-7 - \lambda^2| = |-7 - 1| = |-8|$$

The final result is **8**.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + 2 \hat{j} + λ \hat{k}$ , $b = 3 \hat{i} - 5 \hat{j} - λ \hat{k}$ , $a \cdot c = 7$ , $2 b \cdot c + 43 = 0$ , $a \times c = b \times c$ . Then $∣ a \cdot b ∣$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors

The Cross Product Condition

Establishing Collinearity

Calculating $a - b$

The First Dot Product Condition

The Second Dot Product Condition

Magnitudes and Dot Product

Forming the First Equation

Forming the Second Equation

Solving for $μ$ and $λ^{2}$

Final Calculation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Power of Rearrangement

Bridging the Gap with Dot Products

The Final Convergence

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^+2j^​+λk^, b=3i^−5j^​−λk^, a⋅c=7, 2b⋅c+43=0, a×c=b×c. Then ∣a⋅b∣ is equal to