If the projection of the vector i + 2j + k on the sum of the two vectors 2i + 4j - 5k and -λi + 2j +

Q: Wait, why is the projection formula $\frac{\vec{a} \cdot \vec{s}}{|\vec{s}|}$ and not just the dot product?

That is a fantastic question! The dot product $\vec{a} \cdot \vec{s}$ actually gives you the product of the magnitudes and the cosine of the angle between them, which is $|\vec{a}||\vec{s}| \cos\theta$. The projection of $\vec{a}$ onto $\vec{s}$ is simply $|\vec{a}| \cos\theta$. If you divide the dot product by the magnitude of the vector you are projecting onto ($|\vec{s}|$), you are left with exactly $|\vec{a}| \cos\theta$. Think of it as normalizing the base vector to unit length so you only measure the shadow's length.

Q: Can the projection be negative? What if the dot product is negative?

Absolutely! The projection is a scalar value, not a distance. If the angle between the two vectors is obtuse (greater than $90^\circ$), the cosine of that angle is negative, making the projection negative. This just means the shadow falls in the opposite direction of the vector $\vec{s}$. In this specific problem, the result is $1$, which is positive, so we don't need to worry about the sign, but always keep that in mind for other problems!

Q: I ended up with a quadratic equation, but the solution says it becomes linear. Did I mess up the algebra?

Don't panic! When you square both sides, you get $\lambda^2$ terms on both sides of the equation. If you expand carefully, you will see that they are identical. When you subtract $\lambda^2$ from both sides, they vanish, leaving you with a beautiful, simple linear equation. It is a common 'trap' to think you've made a mistake, but it is actually a sign that you are on the right track.

Question

If the projection of the vector $\hat{i} + 2\hat{j} + \hat{k}$ on the sum of the two vectors $2\hat{i} + 4\hat{j} - 5\hat{k}$ and $-\lambda\hat{i} + 2\hat{j} + 3\hat{k}$ is 1, then $\lambda$ is equal to .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You have a vector $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ floating in front of you.

A second vector, $\vec{s}$, is defined as the sum of two other vectors: $\vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{c} = -\lambda\hat{i} + 2\hat{j} + 3\hat{k}$. We must find the value of $\lambda$ such that the projection of $\vec{a}$ onto $\vec{s}$ is exactly $1$.

The projection is essentially the shadow of $\vec{a}$ cast onto the line of $\vec{s}$. When we say the projection is $1$, we are stating that the length of the shadow cast by $\vec{a}$ onto the direction of $\vec{s}$ is exactly $1$ unit.

### The Algebraic Foundation

First, we determine our base vector $\vec{s}$ by summing the components of $\vec{b}$ and $\vec{c}$:

$$\vec{s} = (2 - \lambda)\hat{i} + 6\hat{j} - 2\hat{k}$$

Next, we invoke the fundamental formula for the scalar projection of $\vec{a}$ onto $\vec{s}$:

$$\frac{\vec{a} \cdot \vec{s}}{|\vec{s}|} = 1$$

We calculate the dot product $\vec{a} \cdot \vec{s}$ by multiplying corresponding components:

$$\vec{a} \cdot \vec{s} = (1)(2 - \lambda) + (2)(6) + (1)(-2)$$

Simplifying this expression, we get $2 - \lambda + 12 - 2$, which elegantly reduces to:

$$\vec{a} \cdot \vec{s} = 12 - \lambda$$

Now, we find the magnitude $|\vec{s}|$:

$$|\vec{s}| = \sqrt{(2 - \lambda)^2 + 6^2 + (-2)^2}$$

Expanding the square, we obtain:

$$|\vec{s}| = \sqrt{\lambda^2 - 4\lambda + 4 + 36 + 4} = \sqrt{\lambda^2 - 4\lambda + 44}$$

### The Moment of Truth

We substitute these results back into our projection equation:

$$\frac{12 - \lambda}{\sqrt{\lambda^2 - 4\lambda + 44}} = 1$$

To solve for $\lambda$, we multiply both sides by the denominator and square both sides:

$$(12 - \lambda)^2 = \lambda^2 - 4\lambda + 44$$

Expanding the left side using the identity $(a - b)^2 = a^2 - 2ab + b^2$, we obtain:

$$144 - 24\lambda + \lambda^2 = \lambda^2 - 4\lambda + 44$$

The $\lambda^2$ terms on both sides cancel out perfectly, leaving us with a linear equation:

$$144 - 24\lambda = -4\lambda + 44$$

Rearranging the terms to isolate $\lambda$:

$$100 = 20\lambda$$

This leads us directly to the final result:

**$\lambda = 5$**

Animated Solution for Mathematics - Vector Algebra: If the projection of the vector $\hat{i} + 2 \hat{j} + \hat{k}$ on the sum of the two vectors $2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $- λ \hat{i} + 2 \hat{j} + 3 \hat{k}$ is 1, then $λ$ is equal to .

Enter Numerical Value:

Visualized Solution

Identify the Primary Vector $a$

Define Vectors $b$ and $c$

Calculate the Sum Vector $s$

The Projection Concept

Calculate Dot Product $a \cdot s$

Calculate Magnitude $∣ s ∣$

Set up the Equation

Square Both Sides

Simplify the Equation

Final Answer

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Algebraic Foundation

The Moment of Truth

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

If $x$ and $y$ be two non-zero vectors such that $∣ x + y ∣ = ∣ x ∣$ and $2 x + λ y$ is perpendicular to $y$ , then the value of $λ$ is

Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}, b = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ and $c = 3 \hat{i} - \hat{j} + λ \hat{k}$ be three vectors. Let $r$ be a unit vector along $b + c$ . If $r \cdot a = 3$ , then $3 λ$ is equal to:

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

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

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

Calculate the Sum Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s