Let a = 2i + j - 2k, b = i + j and c = a × b. Let d be a vector such that |d - a| = √11, |c × d| = 3

Q: Why did we square the magnitude $|\vec{d} - \vec{a}| = \sqrt{11}$?

Great question! When we have an equation involving the magnitude of a vector difference, it is almost always a signal to square both sides. This transforms the geometric magnitude into an algebraic dot product expression: $|\vec{d} - \vec{a}|^2 = |\vec{d}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{d})$. This is the key that unlocks the dot product we need to find.

Q: Is the dot product $\vec{a} \cdot \vec{d}$ always zero when the magnitudes are specific values?

Not necessarily! In this specific problem, the numbers worked out such that the dot product is zero, which implies the vectors $\vec{a}$ and $\vec{d}$ are orthogonal. This is a beautiful result, but it is a consequence of the specific values given in the problem, not a general rule for all vectors.

Q: Can we find the vector $\vec{d}$ itself?

With the information provided, we can find the magnitude of $\vec{d}$ and its dot product with $\vec{a}$, but we cannot uniquely determine the vector $\vec{d}$ itself. There are infinitely many vectors $\vec{d}$ that satisfy these conditions, forming a cone-like structure in 3D space. However, for the purpose of finding $\vec{a} \cdot \vec{d}$, we have all the necessary information!

Question

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

Options:

(A) 3

(B) 11

(C) 0

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two vectors in 3D space: $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$. These vectors define a plane, and we first determine the normal vector $\vec{c} = \vec{a} 	imes \vec{b}$ using the determinant expansion:

$$
\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix}
$$

Expanding this determinant, we obtain:

$$
\vec{c} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}
$$

Next, we calculate the magnitudes of these vectors. For $\vec{a}$:

$$
|\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3
$$

For the normal vector $\vec{c}$:

$$
|\vec{c}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3
$$

### Decoding the Mystery Vector $\vec{d}$

We are introduced to a vector $\vec{d}$ where $|\vec{c} 	imes \vec{d}| = 3$ and the angle between $\vec{c}$ and $\vec{d}$ is $	heta = \frac{\pi}{4}$. Using the geometric definition of the cross product, $|\vec{c} 	imes \vec{d}| = |\vec{c}| |\vec{d}| \sin(	heta)$, we substitute the known values:

$$
3 = 3 \cdot |\vec{d}| \cdot \sin\left(\frac{\pi}{4}ight)
$$

Since $\sin\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$, the equation becomes:

$$
3 = 3 \cdot |\vec{d}| \cdot \frac{1}{\sqrt{2}}
$$

Solving for the magnitude, we find $|\vec{d}| = \sqrt{2}$.

### The Final Climax: The Dot Product

We are given the condition $|\vec{d} - \vec{a}| = \sqrt{11}$. To extract the dot product $\vec{a} \cdot \vec{d}$, we square both sides of the equation:

$$
|\vec{d} - \vec{a}|^2 = 11
$$

Expanding the dot product identity $|\vec{d} - \vec{a}|^2 = |\vec{d}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{d})$, we substitute the known magnitudes $|\vec{d}|^2 = 2$ and $|\vec{a}|^2 = 9$:

$$
2 + 9 - 2(\vec{a} \cdot \vec{d}) = 11
$$

This simplifies to $11 - 2(\vec{a} \cdot \vec{d}) = 11$. Subtracting 11 from both sides yields $-2(\vec{a} \cdot \vec{d}) = 0$.

Therefore, the final result is:

**$\vec{a} \cdot \vec{d} = 0$**

This indicates that the vectors $\vec{a}$ and $\vec{d}$ are **orthogonal**.

Animated Solution for Mathematics - Vector Algebra: 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

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Visualized Solution

Given Vectors $a$ and $b$

Setting up the Cross Product

Calculating Vector $c$

Magnitudes of $a$ and $c$

Cross Product Condition for $d$

Finding Magnitude of $d$

The Difference Vector Condition

Expanding the Squared Magnitude

Substituting Known Magnitudes

Solving for $a \cdot d$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Decoding the Mystery Vector $d$

The Final Climax

Similar Questions

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Select Answer:

Visualized Solution

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

Setting up the Cross Product

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

Magnitudes of .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; } c

Cross Product Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Finding Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

The Difference Vector Condition

Expanding the Squared Magnitude

Substituting Known Magnitudes

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

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Decoding the Mystery Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

The Final Climax

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