Let a = 3i + j - k and c = 2i - 3j + 3k. If b is a vector such that a = b × c and |b|² = 50, then |7

Q: Wait, why do we have a plus-minus sign for $\cos \theta$?

Great catch! When we find $\sin^2 \theta = \frac{1}{100}$, we know $\cos^2 \theta = 1 - \sin^2 \theta = \frac{99}{100}$. Taking the square root gives $\cos \theta = \pm \frac{\sqrt{99}}{10}$. Since the problem doesn't specify the orientation of $\vec{b}$ relative to $\vec{c}$, both acute and obtuse angles are mathematically possible. However, as you will see, the final expression $|72 - |\vec{b} + \vec{c}|^2|$ will yield the same result regardless of the sign.

Q: Why can't we just find the components of $\vec{b}$ first?

That is a natural instinct, but it is a trap! We are given $\vec{a} = \vec{b} \times \vec{c}$, which is one vector equation with three unknowns (the components of $\vec{b}$). This is an underdetermined system. We cannot uniquely solve for $\vec{b}$ because there are infinitely many vectors $\vec{b}$ that satisfy this cross product relation. We must rely on the geometric properties of magnitudes and angles instead.

Q: Is there a trick to the final cancellation?

It might look like a coincidence, but it is the beauty of the problem's design. When you expand $|\vec{b} + \vec{c}|^2$, you get $50 + 22 + 2|\vec{b}||\vec{c}|\cos \theta$. The $50 + 22$ sums to $72$. The dot product term simplifies to $\pm 66$. So, $|\vec{b} + \vec{c}|^2$ becomes $72 \pm 66$. When you subtract this from $72$ inside the absolute value, the $72$ cancels out, leaving you with $|\mp 66|$, which is simply $66$.

Question

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

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: A Journey into the Unknown

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are unraveling a mystery hidden within the structure of vectors.

We are given two vectors, $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} - 3\hat{j} + 3\hat{k}$, and a cryptic relationship: $\vec{a} = \vec{b} 	imes \vec{c}$. We are also told that $|\vec{b}|^2 = 50$.

Our mission is to find the value of $|72 - |\vec{b} + \vec{c}|^2|$. Let us embark on this journey step by step.

### Phase 1: The Building Blocks

Before we dive into the deep waters of vector algebra, we must gather our tools. We start by calculating the squared magnitudes of the vectors we know.

For $\vec{a}$, the squared magnitude is:
$$|\vec{a}|^2 = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11$$

For $\vec{c}$, it is:
$$|\vec{c}|^2 = 2^2 + (-3)^2 + 3^2 = 4 + 9 + 9 = 22$$

And we are handed the squared magnitude of $\vec{b}$ on a silver platter: $|\vec{b}|^2 = 50$. These numbers—$11$, $22$, and $50$—are the keys to our kingdom.

### Phase 2: The Cross Product Mystery

Now, consider the relation $\vec{a} = \vec{b} 	imes \vec{c}$. Geometrically, this tells us that $\vec{a}$ is the normal vector to the plane defined by $\vec{b}$ and $\vec{c}$.

Algebraically, it gives us a powerful tool: the magnitude of the cross product. We know that $|\vec{a}|^2 = |\vec{b} 	imes \vec{c}|^2$.

Using the fundamental definition of the cross product magnitude, this becomes:
$$|\vec{a}|^2 = |\vec{b}|^2 |\vec{c}|^2 \sin^2 	heta$$
where $	heta$ is the angle between $\vec{b}$ and $\vec{c}$.

Substituting our known values, we get:
$$11 = 50 	imes 22 	imes \sin^2 	heta$$

Simplifying this, $11 = 1100 \sin^2 	heta$, which leads us to the elegant result:
$$\sin^2 	heta = \frac{11}{1100} = \frac{1}{100}$$

### Phase 3: The Dot Product Bridge

With $\sin^2 	heta$ in our grasp, finding $\cos 	heta$ is just a step away. Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$, we find:
$$\cos^2 	heta = 1 - \frac{1}{100} = \frac{99}{100}$$

Taking the square root, we get $\cos 	heta = \pm \frac{\sqrt{99}}{10} = \pm \frac{3\sqrt{11}}{10}$. This plus-minus sign is a reminder that the geometry of the vectors allows for two possible configurations.

Now, we turn our attention to the expression $|\vec{b} + \vec{c}|^2$. Expanding this using the dot product property, we get:
$$|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{b} \cdot \vec{c})$$

Since $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}| \cos 	heta$, we have all the components ready to assemble.

### Phase 4: The Grand Finale

Let us substitute our values into the expansion:
$$|\vec{b} + \vec{c}|^2 = 50 + 22 + 2\sqrt{50}\sqrt{22} \left( \pm \frac{3\sqrt{11}}{10} ight)$$

Breaking down the roots, $\sqrt{50} = 5\sqrt{2}$ and $\sqrt{22} = \sqrt{2}\sqrt{11}$. The expression becomes:
$$72 \pm 2(5\sqrt{2})(\sqrt{2}\sqrt{11}) \left( \frac{3\sqrt{11}}{10} ight)$$

Simplifying the terms, we see that $2 	imes 5 	imes 2 	imes 11 	imes \frac{3}{10} = 66$. Thus, $|\vec{b} + \vec{c}|^2 = 72 \pm 66$.

Finally, we evaluate the expression $|72 - |\vec{b} + \vec{c}|^2|$. Substituting our result:
$$|72 - (72 \pm 66)| = |\mp 66| = 66$$

The $72$ cancels out, leaving us with the beautiful, absolute value of **$66$**. We have conquered the problem!

Animated Solution for Mathematics - Vector Algebra: 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

Given Vectors and Magnitudes

Cross Product Relation

Magnitude of Cross Product

Solving for $sin^{2} θ$

Finding $cos θ$

Expanding $∣ b + c ∣^{2}$

Substituting Values into Expansion

Simplifying the Dot Product Term

Final Calculation

The Sigma Insight: Vector (Cross) Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $λ \in Z$ , $a = λ \hat{i} + \hat{j} - \hat{k}$ and $b = 3 \hat{i} - \hat{j} + 2 \hat{k}$ . Let $c$ be a vector such that $(a + b + c) \times c = 0$ , $a \cdot c = - 17$ and $b \cdot c = - 20$ . Then $∣ c \times (λ \hat{i} + \hat{j} + \hat{k}) ∣^{2}$ 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 $a = \hat{i} - 3 \hat{j} + 7 \hat{k}$ , $b = 2 \hat{i} - \hat{j} + \hat{k}$ 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 $a$ and $b$ be two vectors such that $∣ a + b ∣^{2} = ∣ a ∣^{2} + 2∣ b ∣^{2}, a \cdot b = 3$ and $∣ a \times b ∣^{2} = 75$ . Then $∣ a ∣^{2}$ 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 $a = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ , $b = \hat{i} - 2 \hat{j} - 2 \hat{k}$ and $c = - \hat{i} + 4 \hat{j} + 3 \hat{k}$ . If $d$ is a vector perpendicular to both $b$ and $c$ , and $a \cdot d = 18$ , then $∣ a \times d ∣^{2}$ 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 $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ 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 $∣ (a \times b) \times c ∣ =$

Let $a = 2 \hat{i} - 3 \hat{j} + \hat{k}, b = 3 \hat{i} + 2 \hat{j} + 5 \hat{k}$ and a vector $c$ be such that $(a - c) \times b = - 18 \hat{i} - 3 \hat{j} + 12 \hat{k}$ and $a \cdot c = 3$ . If $b \times c = d$ , then $∣ a \cdot d ∣$ is equal to :

Let $a = \hat{i} + \hat{j} + \hat{k}, c = \hat{j} - \hat{k}$ and a vector $b$ be such that $a \times b = c$ and $a \cdot b = 3$ . Then $∣ b ∣$ equals :

Enter Numerical Value:

Visualized Solution

Given Vectors and Magnitudes

Cross Product Relation

Magnitude of Cross Product

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2θ

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

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣b+c∣2

Substituting Values into Expansion

Simplifying the Dot Product Term

Final Calculation

The Sigma Insight: Vector (Cross) Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let λ∈Z, a=λi^+j^​−k^ and b=3i^−j^​+2k^. Let c be a vector such that (a+b+c)×c=0, a⋅c=−17 and b⋅c=−20. Then ∣c×(λi^+j^​+k^)∣2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=−i^+2j^​+2k^,b=8i^+7j^​−3k^ and c be vector such that a×c=b. If c⋅(i^+j^​+k^)=4, then ∣a+c∣2 is equal to :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a and b be two vectors such that ∣a+b∣2=∣a∣2+2∣b∣2,a⋅b=3 and ∣a×b∣2=75. Then ∣a∣2 is equal to

.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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^−5j^​+5k^ and b=i^−j^​+3k^. If c is a vector such that 2(a×c)+3(b×c)=0 and (a−b)⋅c=−97, then ∣c×k^∣2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^+j^​−2k^ and b=i^+j^​. If c is a vector such that a⋅c=∣c∣,∣c−a∣=22​ and the angle between (a×b) and c is 30∘, then ∣(a×b)×c∣=

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^+j^​+k^,c=j^​−k^ and a vector b be such that a×b=c and a⋅b=3. Then ∣b∣ equals :