Let a=i+j+k, b=2i+2j+k and d=a×b If c is a vector such that a·c=|c|, |c-2a|²=8 and the angle between

Q: Wait, why do we expand $|\vec{c} - 2\vec{a}|^2$ using dot products instead of just treating it like a scalar $(c-2a)^2$?

That is a fantastic question! In vector algebra, we cannot treat vectors like scalars because they have both magnitude and direction. The square of a vector is defined as the dot product of the vector with itself: $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$. Therefore, $|\vec{c} - 2\vec{a}|^2$ is actually $(\vec{c} - 2\vec{a}) \cdot (\vec{c} - 2\vec{a})$. When you distribute this dot product, you get $|\vec{c}|^2 - 4(\vec{a} \cdot \vec{c}) + 4|\vec{a}|^2$. This is the only mathematically rigorous way to handle the magnitude of a vector sum or difference.

Q: Why are there two possible values for $\vec{c}$? Did I miss a condition in the problem?

You didn't miss anything! The constraints provided—the magnitude of $\vec{c}$, its dot product with $\vec{a}$, and its dot product with $\vec{d}$—define a system of equations that can have multiple solutions in 3D space. Geometrically, this corresponds to the intersection of a sphere and two planes. It just so happens that in this specific problem, both potential vectors $\vec{c}$ lead to the same final result for the expression $|10 - 3\vec{b} \cdot \vec{c}| + |\vec{d} \times \vec{c}|^2$. It is a beautiful feature of the problem's design.

Q: Is there a shortcut for calculating $|\vec{d} \times \vec{c}|^2$ without finding the components of $\vec{c}$?

Yes, there is! You can use the definition $|\vec{d} \times \vec{c}| = |\vec{d}| |\vec{c}| \sin(\theta)$. Since you are given the angle $\theta = \frac{\pi}{4}$ between $\vec{c}$ and $\vec{d}$, and you can easily calculate $|\vec{d}|$ and $|\vec{c}|$, you can find the magnitude of the cross product directly. Squaring this value gives you the term $|\vec{d} \times \vec{c}|^2$ immediately, saving you from having to perform the cross product calculation with unknown variables.

Question

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}.$, $\vec{b}=2\hat{i}+2\hat{j}+\hat{k}$ and $\vec{d}=\vec{a}	imes\vec{b}$ If $\vec{c}$ is a vector such that $\vec{a}\cdot\vec{c}=|\vec{c}|$, $|\vec{c}-2\vec{a}|^{2}=8$ and the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$ then $|10-3\vec{b}\cdot\vec{c}|+|\vec{d}	imes\vec{c}|^{2}$ is equal to

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: A Journey into 3D Space

Welcome, future engineer. Today, we are going to dissect a problem that is not merely a collection of algebraic manipulations, but a beautiful exploration of three-dimensional geometry.

When you look at vectors like $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}$, do not just see numbers. See arrows in space, defining planes, angles, and magnitudes. Our goal is to find a mysterious vector $\vec{c}$ that satisfies a set of rigid constraints.

Let us embark on this journey.

### Phase 1: Unlocking the Magnitude

We begin with the condition $|\vec{c} - 2\vec{a}|^2 = 8$. Many students feel the urge to treat this like a simple scalar identity, but remember: vectors are not scalars.

We must expand this using the dot product property: $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 - 2\vec{u} \cdot \vec{v} + |\vec{v}|^2$. Applying this to our equation, we get:

$$|\vec{c}|^2 - 4(\vec{a} \cdot \vec{c}) + 4|\vec{a}|^2 = 8$$

We know that $|\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3$. The problem also gifts us the condition $\vec{a} \cdot \vec{c} = |\vec{c}|$.

Substituting these into our expansion, we transform the equation into a quadratic form:

$$|\vec{c}|^2 - 4|\vec{c}| + 4(3) = 8 \implies |\vec{c}|^2 - 4|\vec{c}| + 12 = 8$$

Rearranging this, we find $|\vec{c}|^2 - 4|\vec{c}| + 4 = 0$, which is the perfect square $(|\vec{c}| - 2)^2 = 0$. Thus, the magnitude of our mystery vector is fixed: $|\vec{c}| = 2$.

We have successfully pinned down the length of $\vec{c}$.

### Phase 2: The Cross Product and the Angle

Next, we turn our attention to $\vec{d} = \vec{a} 	imes \vec{b}$. Using the determinant method, we calculate:

$$\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & 2 & 1 \end{vmatrix} = \hat{i}(1-2) - \hat{j}(1-2) + \hat{k}(2-2) = -\hat{i} + \hat{j}$$

The magnitude $|\vec{d}|$ is $\sqrt{(-1)^2 + 1^2} = \sqrt{2}$. We are told the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$.

This allows us to calculate the cross product magnitude squared directly: $|\vec{d} 	imes \vec{c}|^2 = |\vec{d}|^2 |\vec{c}|^2 \sin^2(\frac{\pi}{4})$. Substituting our values:

$$2 	imes 4 	imes \left( \frac{1}{\sqrt{2}} ight)^2 = 2 	imes 4 	imes \frac{1}{2} = 4$$

We have found one part of our final expression!

### Phase 3: The Two-Path Mystery

Now, we must find $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$. We have two dot product conditions:

1. $\vec{d} \cdot \vec{c} = |\vec{d}| |\vec{c}| \cos(\frac{\pi}{4}) = \sqrt{2} 	imes 2 	imes \frac{1}{\sqrt{2}} = 2$. Since $\vec{d} = -\hat{i} + \hat{j}$, this gives $-x + y = 2$, or $y = x + 2$.

2. $\vec{a} \cdot \vec{c} = 2$. Since $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, this gives $x + y + z = 2$. Substituting $y = x + 2$, we get $x + (x + 2) + z = 2$, which simplifies to $z = -2x$.

Using the magnitude $|\vec{c}|^2 = 4$, we write $x^2 + (x+2)^2 + (-2x)^2 = 4$. Expanding this, we get $x^2 + x^2 + 4x + 4 + 4x^2 = 4$, leading to $6x^2 + 4x = 0$.

This gives us two cases: $x = 0$ or $x = -\frac{2}{3}$.

### The Final Convergence

In Case 1 ($x=0$), $\vec{c} = 2\hat{j}$. Then $\vec{b} \cdot \vec{c} = (2\hat{i} + 2\hat{j} + \hat{k}) \cdot (2\hat{j}) = 4$. The expression becomes $|10 - 3(4)| + 4 = |-2| + 4 = 6$.

In Case 2 ($x=-\frac{2}{3}$), $\vec{c} = -\frac{2}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{4}{3}\hat{k}$. Then $\vec{b} \cdot \vec{c} = 2(-\frac{2}{3}) + 2(\frac{4}{3}) + 1(\frac{4}{3}) = \frac{8}{3}$. The expression becomes $|10 - 3(\frac{8}{3})| + 4 = |10 - 8| + 4 = 6$.

In both cases, the result is **6**. The beauty of mathematics is that even when the path splits, the truth remains singular. You have conquered the problem.

Enter Numerical Value:

Visualized Solution

Visualizing the Given Vectors

Magnitude of Vector $a$

Expanding the Magnitude Equation

Substitution and Simplification

Finding the Magnitude of $c$

Calculating Vector $d$

Magnitude of Vector $d$

Cross Product Magnitude $∣ d \times c ∣^{2}$

Dot Product $d \cdot c$

Setting up Equations for Vector $c$

Solving for Components of $c$

Calculating Dot Product $b \cdot c$

Final Evaluation

The Sigma Insight: Scalar Triple Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

The Final Convergence

Similar Questions

If $a = α \hat{i} + β \hat{j} + 3 \hat{k}$ , $b = β \hat{i} - α \hat{j} - \hat{k}$ and $c = \hat{i} - 2 \hat{j} - \hat{k}$ such that $a \cdot b = 1$ and $b \cdot c = - 3$ , then $\frac{1}{3} ((a \times b) \cdot c)$ is equal to

Let $c$ be a vector perpendicular to the vectors $a = \hat{i} + \hat{j} - \hat{k}$ and $b = \hat{i} + 2 \hat{j} + \hat{k}$ . If $c \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 8$ then the value of $c \cdot (a \times b)$ is equal to

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

Let $a = \hat{i} - \hat{j}$ , $b = \hat{j} - \hat{k}$ , $c = \hat{k} - \hat{i}$ . If $d$ is a unit vector such that $a \cdot d = 0 = [b c d]$ , then $d$ equals

Let $a = \hat{i} + \hat{j} + \hat{k}$ , $b = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $c = x \hat{i} + 2 \hat{j} + 3 \hat{k}, x \in R$ . If $d$ is the unit vector in the direction of $b + c$ such that $a \cdot d = 1$ , then $(a \times b) \cdot c$ is equal to

Let a vector $a$ be coplanar with vectors $b = 2 \hat{i} + \hat{j} + \hat{k}$ and $c = \hat{i} - \hat{j} + \hat{k}$ . If $a$ is perpendicular to $d = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ , and $∣ a ∣ = 10$ . Then a possible value of $[a b c] + [a b d] + [a c d]$ is equal to:

Let $a$ and $b$ be two vectors. Let $∣ a ∣ = 1, ∣ b ∣ = 4$ and $a \cdot b = 2$ . If $c = (2 a \times b) - 3 b$ , then the value of $b \cdot c$ is

Let $a, b$ and $c$ be three non-coplanar unit vectors such that the angle between every pair of them is $\frac{π}{3}$ . If $a \times b + b \times c = p a + q b + r c$ , where $p, q$ and $r$ are scalars, then the value of $\frac{p ^{2} + 2 q ^{2} + r ^{2}}{q ^{2}}$ is .........

Let $a = \hat{i} - α \hat{j} + β \hat{k}, b = 3 \hat{i} + β \hat{j} - α \hat{k}$ and $c = - α \hat{i} - 2 \hat{j} + \hat{k}$ , where $α$ and $β$ are integers. If $a \cdot b = - 1$ and $b \cdot c = 10$ , then $(a \times b) \cdot c$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Given Vectors

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

Expanding the Magnitude Equation

Substitution and Simplification

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

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

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

Cross Product Magnitude .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣d×c∣2

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

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

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

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