Let b = 4i + 3j and c be two vectors perpendicular to each other in the xy-plane. All vectors in the

Q: Wait, why do we need to find $\vec{c}$ explicitly? Can't we just use the dot product properties?

That is a brilliant observation! While you could theoretically work with the abstract properties of the dot product, finding $\vec{c}$ explicitly makes the system of equations concrete and much easier to solve. By identifying $\vec{c} = 3\hat{i} - 4\hat{j}$, you transform an abstract vector problem into a standard system of linear equations, which is the most reliable path to the correct answer.

Q: Is it possible for the projection to be negative? What does that mean?

Absolutely! A negative projection simply means that the vector $\vec{a}$ has a component pointing in the opposite direction of the vector it is being projected onto. In this specific problem, the projections are positive, so $\vec{a}$ makes an acute angle with both $\vec{b}$ and $\vec{c}$. Always trust the math—the sign of the dot product will tell you the directionality automatically.

Q: Why can't we just use the formula $\vec{a} = (proj_b \vec{a}) \hat{b} + (proj_c \vec{a}) \hat{c}$?

You actually can! That is a very elegant way to solve it. If you calculate the unit vectors $\hat{b}$ and $\hat{c}$ and multiply them by their respective projections (1 and 2), you will get the exact same vector $\vec{a}$. It is a fantastic way to verify your answer. The method used in the solution is just a more algebraic approach that many students find more intuitive for solving systems.

Question

Let $\vec{b} = 4\hat{i} + 3\hat{j}$ and $\vec{c}$ be two vectors perpendicular to each other in the $xy$-plane. All vectors in the same plane having projections 1 and 2 along $\vec{b}$ and $\vec{c}$, respectively, are given by .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to the beautiful world of vectors. Today, we are not just solving a problem; we are learning to see the $xy$-plane through a different lens.

We are given a vector $\vec{b} = 4\hat{i} + 3\hat{j}$. Before we do anything else, let us ground ourselves in its physical reality.

The magnitude of this vector is:
$$|\vec{b}| = \sqrt{4^2 + 3^2} = 5$$

This is our anchor. Now, we are introduced to a mysterious partner, vector $\vec{c}$, which is perpendicular to $\vec{b}$. In the language of linear algebra, $\vec{b}$ and $\vec{c}$ are orthogonal; they are the foundation of a new coordinate system.

### The Perpendicular Partner

If $\vec{b} = 4\hat{i} + 3\hat{j}$ and $\vec{c} = \alpha\hat{i} + \beta\hat{j}$ are perpendicular, their dot product must vanish: $\vec{b} \cdot \vec{c} = 0$. This gives us the elegant constraint:
$$4\alpha + 3\beta = 0$$

To find a vector $\vec{c}$ that satisfies this, we can simply swap the coefficients and negate one: $\vec{c} = 3\hat{i} - 4\hat{j}$. Notice the beauty here? The magnitude of $\vec{c}$ is also:
$$|\vec{c}| = \sqrt{3^2 + (-4)^2} = 5$$

We have successfully constructed an orthogonal basis.

### The Unknown Hero

Now, let us define our target, the unknown vector $\vec{a} = x\hat{i} + y\hat{j}$. We are looking for the coordinates $(x, y)$ that define this vector.

The problem gives us two clues: the projection of $\vec{a}$ along $\vec{b}$ is $1$, and the projection along $\vec{c}$ is $2$. The projection of $\vec{a}$ onto any vector $\vec{v}$ is defined as:
$$	ext{proj}_{\vec{v}} \vec{a} = \frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}$$

This is the bridge between our unknown components and the given values.

### The Projection Bridge

Let us apply this to $\vec{b}$:
$$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 1$$

Substituting our components, we get:
$$\frac{4x + 3y}{5} = 1 \Rightarrow 4x + 3y = 5$$

This is our first constraint. Now, for $\vec{c}$:
$$\frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = 2$$

Substituting, we get:
$$\frac{3x - 4y}{5} = 2 \Rightarrow 3x - 4y = 10$$

We have arrived at a system of two linear equations.

### The Final Resolution

We have the system:
$$4x + 3y = 5$$
$$3x - 4y = 10$$

To solve this, we use the elimination method. Multiply the first by $4$ and the second by $3$:
$$16x + 12y = 20$$
$$9x - 12y = 30$$

Adding these equations, the $y$ terms vanish, leaving $25x = 50$, so $x = 2$. Substituting $x=2$ back into the first equation:
$$4(2) + 3y = 5 \Rightarrow 3y = -3 \Rightarrow y = -1$$

Our vector is **$\vec{a} = 2\hat{i} - \hat{j}$**. It is a simple, elegant result. You have successfully navigated the geometry of projections.

Animated Solution for Mathematics - Vector Algebra: Let $b = 4 \hat{i} + 3 \hat{j}$ and $c$ be two vectors perpendicular to each other in the $x y$ -plane. All vectors in the same plane having projections 1 and 2 along $b$ and $c$ , respectively, are given by .........

Visualized Solution

Visualizing Vector $b$

Defining Perpendicular Vector $c$

Finding Components of $c$

Introducing the Unknown Vector $a$

Projection of $a$ on $b$

Projection of $a$ on $c$

Solving the System of Equations

Finding the $x$ -component

Finding the $y$ -component

The Final Vector

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Perpendicular Partner

The Unknown Hero

The Projection Bridge

The Final Resolution

Similar Questions

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 $a = \hat{i} + \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ 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 $a = 2 \hat{i} - \hat{j} + \hat{k}$ , $b = \hat{i} + 2 \hat{j} - \hat{k}$ 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 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 $u, v, w$ 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

The vector(s) which is/are coplanar with vectors $\hat{i} + \hat{j} + 2 \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ , and perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$ is/are

The vector $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$ is

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 :-

Visualized Solution

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

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

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

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

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

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