Let a = 2i - j + 2k and b = i + 2j - k. Let a vector v be in the plane containing a and b. If v is p

Q: Wait, why can't we just find the normal vector $\vec{n} = \vec{a} \times \vec{b}$ and say $\vec{v} = x\vec{a} + y\vec{b}$?

You absolutely can! That is a perfectly valid approach. However, using the vector triple product $\vec{v} = \lambda [\vec{c} \times (\vec{a} \times \vec{b})]$ is often faster and more elegant because it directly incorporates the perpendicularity condition with $\vec{c}$ without needing to solve a system of linear equations for $x$ and $y$. Both paths lead to the same destination, but the VTP path is a shortcut through the woods.

Q: Why is $\vec{v}$ parallel to $\vec{c} \times (\vec{a} \times \vec{b})$?

Think of it this way: $\vec{v}$ is perpendicular to $\vec{c}$ (given) and $\vec{v}$ is perpendicular to the plane of $\vec{a}$ and $\vec{b}$ (because it lies in that plane). The normal to the plane is $\vec{a} \times \vec{b}$. If a vector is perpendicular to two other vectors, it must be parallel to their cross product. Therefore, $\vec{v} \parallel \vec{c} \times (\vec{a} \times \vec{b})$.

Q: Why do we use the projection formula $\frac{\vec{v} \cdot \vec{a}}{|\vec{a}|}$ instead of just $\vec{v} \cdot \vec{a}$?

Great catch! The dot product $\vec{v} \cdot \vec{a}$ gives you the product of the magnitudes and the cosine of the angle between them. The projection of $\vec{v}$ onto $\vec{a}$ is the component of $\vec{v}$ along the direction of $\vec{a}$, which is defined as $\frac{\vec{v} \cdot \vec{a}}{|\vec{a}|}$. The problem specifies the projection is 19 units, which is a scalar length, not the dot product itself.

Question

Let $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$. Let a vector $\vec{v}$ be in the plane containing $\vec{a}$ and $\vec{b}$. If $\vec{v}$ is perpendicular to the vector $3\hat{i} + 2\hat{j} - \hat{k}$ and its projection on $\vec{a}$ is 19 units, then $|2\vec{v}|^2$ is equal to .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty 3D space. You have a flat, infinite sheet of paper floating in front of you, representing a plane defined by two vectors:
$\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.

A mystery vector $\vec{v}$ is trapped on this sheet. Because $\vec{v}$ lies in the plane of $\vec{a}$ and $\vec{b}$, it must be perpendicular to the normal vector of that plane, which is defined by the cross product $\vec{a} 	imes \vec{b}$.

Thus, we establish the constraint: $\vec{v} \perp (\vec{a} 	imes \vec{b})$.

### The Double Perpendicularity

We are also told that $\vec{v}$ is perpendicular to a third vector, $\vec{c} = 3\hat{i} + 2\hat{j} - \hat{k}$.

Since $\vec{v}$ is perpendicular to both $\vec{c}$ and the normal vector $(\vec{a} 	imes \vec{b})$, its direction must be parallel to their cross product. We can define the vector as:
$$\vec{v} = \lambda [\vec{c} 	imes (\vec{a} 	imes \vec{b})]$$
where $\lambda$ is a scalar constant.

### The Elegance of the Vector Triple Product

To simplify the expression $\vec{c} 	imes (\vec{a} 	imes \vec{b})$, we apply the Vector Triple Product (VTP) identity:
$$\vec{A} 	imes (\vec{B} 	imes \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$$

Applying this to our expression, we get:
$$\vec{v} = \lambda [(\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}]$$

Calculating the dot products:
$\vec{c} \cdot \vec{b} = (3)(1) + (2)(2) + (-1)(-1) = 3 + 4 + 1 = 8$
$\vec{c} \cdot \vec{a} = (3)(2) + (2)(-1) + (-1)(2) = 6 - 2 - 2 = 2$

Substituting these values, we find:
$$\vec{v} = \lambda [8\vec{a} - 2\vec{b}]$$

### Bringing it All Together

Substituting the components of $\vec{a}$ and $\vec{b}$ into the equation:
$$\vec{v} = \lambda [8(2\hat{i} - \hat{j} + 2\hat{k}) - 2(\hat{i} + 2\hat{j} - \hat{k})]$$
$$\vec{v} = \lambda [16\hat{i} - 8\hat{j} + 16\hat{k} - 2\hat{i} - 4\hat{j} + 2\hat{k}] = \lambda [14\hat{i} - 12\hat{j} + 18\hat{k}]$$

The problem states the projection of $\vec{v}$ on $\vec{a}$ is $19$. Using the formula $\frac{\vec{v} \cdot \vec{a}}{|\vec{a}|} = 19$:
First, $|\vec{a}| = \sqrt{2^2 + (-1)^2 + 2^2} = 3$.
Next, $\vec{v} \cdot \vec{a} = \lambda [14(2) + (-12)(-1) + 18(2)] = \lambda [28 + 12 + 36] = 76\lambda$.

Plugging these into the projection formula:
$$\frac{76\lambda}{3} = 19 \Rightarrow 76\lambda = 57 \Rightarrow \lambda = \frac{57}{76} = \frac{3}{4}$$

### The Final Victory

We now calculate $2\vec{v}$:
$$2\vec{v} = 2 \cdot \frac{3}{4} [14\hat{i} - 12\hat{j} + 18\hat{k}] = \frac{3}{2} [14\hat{i} - 12\hat{j} + 18\hat{k}] = 3 [7\hat{i} - 6\hat{j} + 9\hat{k}]$$

Finally, we compute the magnitude squared $|2\vec{v}|^2$:
$$|2\vec{v}|^2 = 3^2 [7^2 + (-6)^2 + 9^2] = 9 [49 + 36 + 81] = 9 [166]$$
$$|2\vec{v}|^2 = \mathbf{1494}$$

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors $a$ and $b$

The Perpendicularity Conditions

Direction of Vector $v$

Vector Triple Product Expansion

Calculating Dot Products

Calculating Dot Products

Expressing $v$ in terms of $λ$

Simplifying Vector $v$

Projection of $v$ on $a$

Magnitude of $a$

Calculating $v \cdot a$

Solving for $λ$

Finding $2 v$

Final Calculation of $∣2 v ∣^{2}$

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Double Perpendicularity

The Elegance of the Vector Triple Product

Bringing it All Together

The Final Victory

Similar Questions

Let $a = \hat{i} - \hat{j}, b = \hat{i} + \hat{j} + \hat{k}$ and $c$ be a vector such that $a \times c + b = 0$ and $a \cdot c = 4$ , then $∣ c ∣^{2}$ is equal to :-

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If $a, b, c$ are three non-zero vectors and $\overset{n}{^}$ is a unit vector perpendicular to $c$ such that $a = α b - \overset{n}{^}, (α \neq = 0)$ and $b \cdot c = 12$ , then $∣ c \times (a \times b) ∣$ is equal to:

The vector $a = - \hat{i} + 2 \hat{j} + \hat{k}$ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is $b$ . Then the projection of $3 a + 2 b$ on $c = 5 \hat{i} + 4 \hat{j} + 3 \hat{k}$ is

Let $x, y$ and $z$ be three vectors each of magnitude $2$ and the angle between each pair of them is $\frac{π}{3}$ . If $a$ is a non-zero vector perpendicular to $x$ and $y \times z$ and $b$ is a non-zero vector perpendicular to $y$ and $z \times x$ , then

Enter Numerical Value:

Visualized Solution

Visualizing the 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

The Perpendicularity Conditions

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

Vector Triple Product Expansion

Calculating Dot Products

Calculating Dot Products

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

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

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

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

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

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

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2v

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣2v∣2

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Double Perpendicularity

The Elegance of the Vector Triple Product

Bringing it All Together

The Final Victory

Similar Questions

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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^+j^​−2k^ and b=i^+j^​. Let c be a vector such that ∣c−a∣=3, ∣(a×b)×c∣=3 and the angle between c and a×b be 30∘. Then a⋅c is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a,b,c are three non-zero vectors and n^ is a unit vector perpendicular to c such that a=αb−n^,(α=0) and b⋅c=12, then ∣c×(a×b)∣ is equal to:

Visualizing the Vectors $a$ and $b$

Direction of Vector $v$

Expressing $v$ in terms of $λ$

Simplifying Vector $v$

Projection of $v$ on $a$

Magnitude of $a$

Calculating $v \cdot a$

Solving for $λ$

Finding $2 v$

Final Calculation of $∣2 v ∣^{2}$

Let $a = \hat{i} - \hat{j}, b = \hat{i} + \hat{j} + \hat{k}$ and $c$ be a vector such that $a \times c + b = 0$ and $a \cdot c = 4$ , then $∣ c ∣^{2}$ is equal to :-

Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ and $b = 2 \hat{i} + \hat{j} - \hat{k}$ . Let $\overset{c}{^}$ be a unit vector in the plane of the vectors $a$ and $b$ and be perpendicular to $a$ . Then such a vector $\overset{c}{^}$ is :

Let $a = 2 \hat{i} - \hat{j} + 5 \hat{k}$ and $b = α \hat{i} + β \hat{j} + 2 \hat{k}$ . If $((a \times b) \times \hat{i}) \cdot \hat{k} = \frac{23}{2}$ , then $∣ b \times 2 \hat{j} ∣$ is equal to

If $a = 2 \hat{i} + \hat{j} + 2 \hat{k}$ , then the value of $∣ \hat{i} \times (a \times \hat{i}) ∣^{2} + ∣ \hat{j} \times (a \times \hat{j}) ∣^{2} + ∣ \hat{k} \times (a \times \hat{k}) ∣^{2}$ is equal to

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . Let $c$ be a vector such that $∣ c - a ∣ = 3$ , $∣ (a \times b) \times c ∣ = 3$ and the angle between $c$ and $a \times b$ be $3 0^{\circ}$ . Then $a \cdot c$ is equal to:

If $a, b, c$ are three non-zero vectors and $\overset{n}{^}$ is a unit vector perpendicular to $c$ such that $a = α b - \overset{n}{^}, (α \neq = 0)$ and $b \cdot c = 12$ , then $∣ c \times (a \times b) ∣$ is equal to: