Let a = 9i - 13j + 25k, b = 3i + 7j - 13k and c = 17i - 2j + k be three given vectors. If r is a vec

Q: Wait, why do we reject the negative root here? Or do we?

In this problem, we are not dealing with a quadratic equation that requires rejecting a root. We are solving a linear equation for the scalar $\lambda$. The value of $\lambda$ is uniquely determined by the dot product condition. There is no ambiguity, so there is no root to reject!

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure vector algebra problem, not a physics problem involving energy. While vectors are used in physics, here we are strictly using the properties of the dot and cross products. Stick to the algebraic definitions of these operations to find the solution.

Q: How do I know $\vec{r} - (\vec{b} + \vec{c})$ is parallel to $\vec{a}$?

This is a fundamental property of the cross product. If $\vec{u} \times \vec{v} = \vec{0}$, then either $\vec{u} = \vec{0}$, $\vec{v} = \vec{0}$, or $\vec{u}$ is parallel to $\vec{v}$. Since $\vec{a}$ is a non-zero vector, the difference vector must be parallel to $\vec{a}$.

Question

Let $\vec{a} = 9\hat{i} - 13\hat{j} + 25\hat{k}, \vec{b} = 3\hat{i} + 7\hat{j} - 13\hat{k}$ and $\vec{c} = 17\hat{i} - 2\hat{j} + \hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} 	imes \vec{a} = (\vec{b} + \vec{c}) 	imes \vec{a}$ and $\vec{r} \cdot (\vec{b} - \vec{c}) = 0$, then $\frac{|593\vec{r} + 67\vec{a}|^2}{(593)^2}$ is equal to _______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional coordinate system, holding three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$. These are not just numbers; they are arrows pointing into the void, defining the structure of the space around you.

Our mission is to find a vector $\vec{r}$ that satisfies two specific constraints. This is a dance of geometry and algebra.

### The Cross Product Insight

We start with the equation $\vec{r} 	imes \vec{a} = (\vec{b} + \vec{c}) 	imes \vec{a}$. If we bring everything to one side, we get:

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

In the language of vectors, if the cross product of two vectors is zero, they must be parallel. This means the vector $(\vec{r} - (\vec{b} + \vec{c}))$ is parallel to $\vec{a}$.

We can capture this relationship with a scalar $\lambda$:

$$\vec{r} = (\vec{b} + \vec{c}) + \lambda\vec{a}$$

### The Dot Product Constraint

We have a second condition: $\vec{r} \cdot (\vec{b} - \vec{c}) = 0$. This tells us that $\vec{r}$ is perpendicular to the vector $(\vec{b} - \vec{c})$.

Substituting our expression for $\vec{r}$ into this condition, we obtain:

$$((\vec{b} + \vec{c}) + \lambda\vec{a}) \cdot (\vec{b} - \vec{c}) = 0$$

Expanding this, we get:

$$(\vec{b} + \vec{c}) \cdot (\vec{b} - \vec{c}) + \lambda\vec{a} \cdot (\vec{b} - \vec{c}) = 0$$

The first part is a classic difference of squares: $|\vec{b}|^2 - |\vec{c}|^2$.

### The Calculation

Given $\vec{b} = 3\hat{i} + 7\hat{j} - 13\hat{k}$, we find:

$$|\vec{b}|^2 = 3^2 + 7^2 + (-13)^2 = 9 + 49 + 169 = 227$$

For $\vec{c} = 17\hat{i} - 2\hat{j} + \hat{k}$, we find:

$$|\vec{c}|^2 = 17^2 + (-2)^2 + 1^2 = 289 + 4 + 1 = 294$$

The difference is $|\vec{b}|^2 - |\vec{c}|^2 = 227 - 294 = -67$.

Next, we calculate $\vec{a} \cdot (\vec{b} - \vec{c})$. First, $\vec{b} - \vec{c} = (3 - 17)\hat{i} + (7 - (-2))\hat{j} + (-13 - 1)\hat{k} = -14\hat{i} + 9\hat{j} - 14\hat{k}$.

Then, $\vec{a} \cdot (\vec{b} - \vec{c}) = 9(-14) - 13(9) + 25(-14) = -126 - 117 - 350 = -593$.

Plugging these into our equation $-67 + \lambda(-593) = 0$, we get:

$$\lambda = -\frac{67}{593}$$

### The Grand Finale

We look at the target expression: $\frac{|593\vec{r} + 67\vec{a}|^2}{(593)^2}$.

Since $\vec{r} = (\vec{b} + \vec{c}) - \frac{67}{593}\vec{a}$, multiplying by $593$ gives $593\vec{r} = 593(\vec{b} + \vec{c}) - 67\vec{a}$.

Rearranging, we find $593\vec{r} + 67\vec{a} = 593(\vec{b} + \vec{c})$. The target expression simplifies to:

$$\frac{|593(\vec{b} + \vec{c})|^2}{(593)^2} = |\vec{b} + \vec{c}|^2$$

Finally, $\vec{b} + \vec{c} = (3+17)\hat{i} + (7-2)\hat{j} + (-13+1)\hat{k} = 20\hat{i} + 5\hat{j} - 12\hat{k}$.

The magnitude squared is:

$$20^2 + 5^2 + (-12)^2 = 400 + 25 + 144 = \mathbf{569}$$

Enter Numerical Value:

Visualized Solution

The Cross Product Relation

The Dot Product Constraint

Calculating Magnitudes

Finding Vector $b - c$

Computing $a \cdot (b - c)$

Solving for $λ$

Simplifying the Target Expression

Calculating $b + c$

Final Result

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Cross Product Insight

The Dot Product Constraint

The Calculation

The Grand Finale

Similar Questions

Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}, b = \hat{i} - \hat{j} + 2 \hat{k}$ and $c = 5 \hat{i} - 3 \hat{j} + 3 \hat{k}$ be there vectors. If $r$ is a vector such that, $r \times b = c \times b$ and $r \cdot a = 0$ . Then $25∣ r ∣^{2}$ is equal to

Let $a = - \hat{i} - \hat{k}$ , $b = - \hat{i} + \hat{j}$ and $c = \hat{i} + 2 \hat{j} + 3 \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times b = c \times b$ and $r \cdot a = 0$ , then the value of $r \cdot b$ is .........

Let $a = 2 \hat{i} - 7 \hat{j} + 5 \hat{k}, b = \hat{i} + \hat{k}$ and $c = \hat{i} + 2 \hat{j} - 3 \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times a = c \times a$ and $r \cdot b = 0$ , then $∣ r ∣$ is equal to:

Let $a = 2 \hat{i} - 5 \hat{j} + 5 \hat{k}$ and $b = \hat{i} - \hat{j} + 3 \hat{k}$ . If $c$ is a vector such that $2 (a \times c) + 3 (b \times c) = 0$ and $(a - b) \cdot c = - 97$ , then $∣ c \times \hat{k} ∣^{2}$ is equal to

Let $a = \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $b = 2 \hat{i} - 3 \hat{j} + 5 \hat{k}$ . If $r \times a = b \times r$ , $r \cdot (α \hat{i} + 2 \hat{j} + \hat{k}) = 3$ and $r \cdot (2 \hat{i} + 5 \hat{j} - α \hat{k}) = - 1$ , $α \in R$ , then the value of $α + ∣ r ∣^{2}$ is equal to :

Let $a = \hat{i} - 3 \hat{j} + 7 \hat{k}$ , $b = 2 \hat{i} - \hat{j} + \hat{k}$ and $c$ be a vector such that $(a + 2 b) \times c = 3 (c \times a)$ . If $a \cdot c = 130$ , then $b \cdot c$ is equal to _______

Let $a = \hat{i} + 2 \hat{j} - \hat{k}, b = \hat{i} - \hat{j}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times a = c \times a$ and $r \cdot b = 0$ , then $r \cdot a$ is equal to

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

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 $a = 3 \hat{i} + \hat{j} - \hat{k}$ and $c = 2 \hat{i} - 3 \hat{j} + 3 \hat{k}$ . If $b$ is a vector such that $a = b \times c$ and $∣ b ∣^{2} = 50$ , then $∣72 - ∣ b + c ∣^{2} ∣$ is equal to

Enter Numerical Value:

Visualized Solution

The Cross Product Relation

The Dot Product Constraint

Calculating Magnitudes

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

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a⋅(b−c)

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

Simplifying the Target Expression

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

Final Result

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Cross Product Insight

The Dot Product Constraint

The Calculation

The Grand Finale

Similar Questions

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Computing $a \cdot (b - c)$

Solving for $λ$

Calculating $b + c$

Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}, b = \hat{i} - \hat{j} + 2 \hat{k}$ and $c = 5 \hat{i} - 3 \hat{j} + 3 \hat{k}$ be there vectors. If $r$ is a vector such that, $r \times b = c \times b$ and $r \cdot a = 0$ . Then $25∣ r ∣^{2}$ is equal to

Let $a = - \hat{i} - \hat{k}$ , $b = - \hat{i} + \hat{j}$ and $c = \hat{i} + 2 \hat{j} + 3 \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times b = c \times b$ and $r \cdot a = 0$ , then the value of $r \cdot b$ is .........

Let $a = 2 \hat{i} - 7 \hat{j} + 5 \hat{k}, b = \hat{i} + \hat{k}$ and $c = \hat{i} + 2 \hat{j} - 3 \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times a = c \times a$ and $r \cdot b = 0$ , then $∣ r ∣$ is equal to:

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