Let x be a vector in the plane containing vectors a = 2i - j + k and b = i + 2j - k. If the vector x

Q: Wait, why can we just write $\vec{x} = \lambda\vec{a} + \mu\vec{b}$? Is that always true?

Great question! This is a fundamental property of vectors in a plane. If $\vec{a}$ and $\vec{b}$ are non-collinear, they form a basis for that plane. Any vector $\vec{x}$ that lies in the same plane can be uniquely expressed as a linear combination of these two basis vectors. It is the geometric equivalent of saying any point on a 2D grid can be reached by moving some distance along the x-axis and some distance along the y-axis.

Q: Why is the projection formula $\frac{\vec{x} \cdot \vec{a}}{|\vec{a}|}$? I thought it was $\frac{\vec{x} \cdot \vec{a}}{|\vec{a}|^2} \vec{a}$?

You are actually both right! The formula $\frac{\vec{x} \cdot \vec{a}}{|\vec{a}|^2} \vec{a}$ gives you the *vector* projection (the shadow itself as a vector). The formula $\frac{\vec{x} \cdot \vec{a}}{|\vec{a}|}$ gives you the *scalar* projection (the length of that shadow). Since the problem gives us a scalar value, we use the scalar projection formula. Always check if the problem asks for the magnitude or the vector itself!

Q: Can we solve this using cross products instead of linear combinations?

Yes, you absolutely can! You could find the normal to the plane using $\vec{n} = \vec{a} \times \vec{b}$, and then use the fact that $\vec{x}$ is perpendicular to both $\vec{n}$ and $\vec{d}$. However, the linear combination method is often faster and less prone to calculation errors in complex 3D problems. Both paths lead to the same beautiful destination.

Question

Let $\vec{x}$ be a vector in the plane containing vectors $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$. If the vector $\vec{x}$ is perpendicular to $(3\hat{i} + 2\hat{j} - \hat{k})$ and its projection on $\vec{a}$ is $\frac{17\sqrt{6}}{2}$, then the value of $|\vec{x}|^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

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

Our goal is to find a mysterious vector $\vec{x}$ that lies on this plane. Because $\vec{x}$ is trapped on this plane, it must be a linear combination of the two vectors that define it:
$$\vec{x} = \lambda\vec{a} + \mu\vec{b}$$

By substituting the components of $\vec{a}$ and $\vec{b}$, we express $\vec{x}$ as:
$$\vec{x} = \lambda(2\hat{i} - \hat{j} + \hat{k}) + \mu(\hat{i} + 2\hat{j} - \hat{k})$$

Grouping the components, we obtain the general form:
$$\vec{x} = (2\lambda + \mu)\hat{i} + (-\lambda + 2\mu)\hat{j} + (\lambda - \mu)\hat{k}$$

### The Constraints of Reality

We are given two clues to pin down the scalars $\lambda$ and $\mu$. First, $\vec{x}$ is perpendicular to $\vec{d} = 3\hat{i} + 2\hat{j} - \hat{k}$, which implies $\vec{x} \cdot \vec{d} = 0$.

Performing the dot product:
$$3(2\lambda + \mu) + 2(-\lambda + 2\mu) - 1(\lambda - \mu) = 0$$

Expanding and simplifying this expression yields our first anchor equation:
$$3\lambda + 8\mu = 0$$

### The Shadow of the Vector

Our second clue involves the scalar projection of $\vec{x}$ onto $\vec{a}$, given as $\frac{17\sqrt{6}}{2}$. The formula for scalar projection is $\frac{\vec{x} \cdot \vec{a}}{|\vec{a}|}$.

First, we calculate the magnitude of $\vec{a}$:
$$|\vec{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$$

Setting up the projection equation:
$$\frac{\vec{x} \cdot \vec{a}}{\sqrt{6}} = \frac{17\sqrt{6}}{2} \Rightarrow \vec{x} \cdot \vec{a} = 51$$

Expanding $\vec{x} \cdot \vec{a}$ using $\vec{x} = \lambda\vec{a} + \mu\vec{b}$, we get $\lambda|\vec{a}|^2 + \mu(\vec{b} \cdot \vec{a}) = 51$. Given $|\vec{a}|^2 = 6$ and $\vec{b} \cdot \vec{a} = -1$, our second equation is:
$$6\lambda - \mu = 51$$

### The Final Resolution

We now solve the system of two linear equations:
1) $3\lambda + 8\mu = 0$
2) $6\lambda - \mu = 51$

From the first equation, $\mu = -\frac{3\lambda}{8}$. Substituting this into the second equation:
$$6\lambda - \left(-\frac{3\lambda}{8}\right) = 51 \Rightarrow \frac{51\lambda}{8} = 51$$

Solving for the scalars, we find $\lambda = 8$ and $\mu = -3$. Plugging these into our general form for $\vec{x}$:
$$\vec{x} = 8(2\hat{i} - \hat{j} + \hat{k}) - 3(\hat{i} + 2\hat{j} - \hat{k}) = 13\hat{i} - 14\hat{j} + 11\hat{k}$$

The final step is to calculate the squared magnitude $|\vec{x}|^2$:
$$|\vec{x}|^2 = 13^2 + (-14)^2 + 11^2 = 169 + 196 + 121 = 486$$

The final result is **486**.

Enter Numerical Value:

Visualized Solution

Defining the Plane of $a$ and $b$

Expressing $x$ in Components

The Perpendicularity Condition

Applying the Dot Product

Projection of $x$ on $a$

Finding $x \cdot a$

Expanding $x \cdot a$

Forming the Second Equation

Solving for $λ$ and $μ$

Finding Vector $x$

Calculating $∣ x ∣^{2}$

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Constraints of Reality

The Shadow of the Vector

The Final Resolution

Similar Questions

Let the vectors $a, b, c$ be such that $∣ a ∣ = 2, ∣ b ∣ = 4$ and $∣ c ∣ = 4$ . If the projection of $b$ on $a$ is equal to the projection of $c$ on $a$ and $b$ is perpendicular to $c$ , then the value of $∣ a + b - c ∣$ is

Let $a = 2 \hat{i} - \hat{j} + \hat{k}$ , $b = \hat{i} + 2 \hat{j} - \hat{k}$ and $c = \hat{i} + \hat{j} - 2 \hat{k}$ be three vectors. A vector in the plane of $b$ and $c$ , whose projection on $a$ is of magnitude $2/3$ , is:

Let $a = 2 \hat{i} + \hat{j} - \hat{k}$ and $b = \hat{i} + 2 \hat{j} + \hat{k}$ be two vectors. Consider a vector $c = α a + β b, α, β \in R$ . If the projection of $c$ on the vector $(a + b)$ is $32$ , then the minimum value of $(c - (a \times b)) \cdot c$ equals

Let $a = 2 \hat{i} - \hat{j} - \hat{k}, b = \hat{i} + 3 \hat{j} - \hat{k}$ and $c = 2 \hat{i} + \hat{j} + 3 \hat{k}$ . Let $v$ be the vector in the plane of the vectors $a$ and $b$ , such that the length of its projection on the vector $c$ is $\frac{1}{14}$ . Then $∣ v ∣$ is equal to

Let $u$ be a vector coplanar with the vectors $a = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $b = \hat{j} + \hat{k}$ . If $u$ is perpendicular to $a$ and $u \cdot b = 24$ , then $∣ u ∣^{2}$ is equal to :

Let $a = \hat{i} + \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three vectors. A vector $v$ in the plane of $a$ and $b$ , whose projection on $c$ is $\frac{1}{3}$ , is given by

Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} + \hat{j} - \hat{k}$ . A vector in the plane of $a$ and $b$ whose projection on $c$ is $\frac{1}{3}$ , is

Let $u, v, w$ be such that $∣ u ∣ = 1, ∣ v ∣ = 2, ∣ w ∣ = 3$ . If the projection $v$ along $u$ is equal to that of $w$ along $u$ and $v, w$ are perpendicular to each other then $∣ u - v + w ∣$ equals

Let $a = \hat{i} - \hat{j} + 2 \hat{k}$ and $b$ be a vector such that $a \times b = 2 \hat{i} - \hat{k}$ and $a \cdot b = 3$ . Then the projection of $b$ on the vector $a - b$ is :-

Enter Numerical Value:

Visualized Solution

Defining the Plane 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; } b

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x in Components

The Perpendicularity Condition

Applying the Dot Product

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

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

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