Let c be a vector perpendicular to the vectors a = i + j - k and b = i + 2j + k. If c . (i + j + 3k)

Q: Wait, why do we assume $\vec{c} = \lambda (\vec{a} \times \vec{b})$? Can't $\vec{c}$ just be any vector perpendicular to the plane?

That is a great question! In 3D space, any vector perpendicular to a plane is necessarily parallel to the normal vector of that plane. Since $\vec{a} \times \vec{b}$ gives us that normal vector, any vector $\vec{c}$ perpendicular to both $\vec{a}$ and $\vec{b}$ must be collinear with $\vec{a} \times \vec{b}$. Therefore, $\vec{c}$ must be a scalar multiple of the cross product.

Q: Is there a way to solve this without using the cross product? Maybe by setting $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$ and solving a system of equations?

Yes, you absolutely could! You would set up two equations from the dot products $\vec{c} \cdot \vec{a} = 0$ and $\vec{c} \cdot \vec{b} = 0$, and use the third given equation $\vec{c} \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 8$. However, that leads to a system of three linear equations with three variables. While valid, it is much more tedious and prone to calculation errors than the elegant cross-product method we used here.

Q: Why does $\vec{c} \cdot (\vec{a} \times \vec{b})$ simplify to $\lambda |\vec{a} \times \vec{b}|^2$?

This is the beauty of vector algebra! We established that $\vec{c} = \lambda (\vec{a} \times \vec{b})$. When you substitute this into the expression $\vec{c} \cdot (\vec{a} \times \vec{b})$, you get $\lambda (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b})$. Since any vector dotted with itself is the square of its magnitude, this becomes $\lambda |\vec{a} \times \vec{b}|^2$. It saves you from having to calculate the components of $\vec{c}$ explicitly!

Question

Let $\vec{c}$ be a vector perpendicular to the vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$. If $\vec{c} \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 8$ then the value of $\vec{c} \cdot (\vec{a} 	imes \vec{b})$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a vast, three-dimensional room. You have two vectors, $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$, anchored at the origin. These two vectors define a unique plane.

The problem asks us to find a vector $\vec{c}$ that is perpendicular to both. Any vector perpendicular to both $\vec{a}$ and $\vec{b}$ must point in the direction of the normal to the plane they span. In the language of vector algebra, this normal is the cross product $\vec{a} 	imes \vec{b}$.

### The Normal Maker

To find this normal, we use the determinant method. We arrange our unit vectors $\hat{i}, \hat{j}, \hat{k}$ in the first row, the components of $\vec{a}$ in the second, and the components of $\vec{b}$ in the third:

$$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -1 \ 1 & 2 & 1 \end{vmatrix}$$

Expanding this, we get $\hat{i}(1 - (-2)) - \hat{j}(1 - (-1)) + \hat{k}(2 - 1)$. This simplifies beautifully to $3\hat{i} - 2\hat{j} + \hat{k}$. This vector is the backbone of our solution.

### The Scalar Multiplier

Since $\vec{c}$ is perpendicular to the plane, it must be parallel to this normal vector. This means $\vec{c}$ is just a scaled version of our cross product. We introduce a scalar $\lambda$ such that $\vec{c} = \lambda (3\hat{i} - 2\hat{j} + \hat{k})$.

We are given one more clue: $\vec{c} \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 8$. This is our key to unlocking $\lambda$. Substituting our expression for $\vec{c}$, we get:

$$\lambda (3\hat{i} - 2\hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 8$$

Calculating the dot product, we have $\lambda (3(1) + (-2)(1) + 1(3)) = 8$, which simplifies to $\lambda (3 - 2 + 3) = 8$, or $4\lambda = 8$. Thus, $\lambda = 2$.

### The Grand Finale

We are asked to find $\vec{c} \cdot (\vec{a} 	imes \vec{b})$. Instead of calculating $\vec{c}$ explicitly, we use the relationship $\vec{c} = 2(\vec{a} 	imes \vec{b})$.

The expression becomes $2(\vec{a} 	imes \vec{b}) \cdot (\vec{a} 	imes \vec{b})$, which is simply $2 |\vec{a} 	imes \vec{b}|^2$. We know the components of $\vec{a} 	imes \vec{b}$ are $(3, -2, 1)$.

The magnitude squared is $3^2 + (-2)^2 + 1^2 = 9 + 4 + 1 = 14$. Finally, $2 	imes 14 = 28$.

The final answer is **28**.

Animated Solution for Mathematics - Vector Algebra: Let $c$ be a vector perpendicular to the vectors $a = \hat{i} + \hat{j} - \hat{k}$ and $b = \hat{i} + 2 \hat{j} + \hat{k}$ . If $c \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 8$ then the value of $c \cdot (a \times b)$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors $a$ and $b$

The Perpendicularity Condition

Setting up the Cross Product

Expanding the Determinant

Result of Cross Product

Defining Vector $c$

Using the Dot Product Condition

Solving for $λ$

Finding the Final Expression

Calculating the Magnitude Squared

The Final Answer

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Normal Maker

The Scalar Multiplier

The Grand Finale

Similar Questions

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

Let $a$ and $b$ be two vectors. Let $∣ a ∣ = 1, ∣ b ∣ = 4$ and $a \cdot b = 2$ . If $c = (2 a \times b) - 3 b$ , then the value of $b \cdot c$ is

If $a = α \hat{i} + β \hat{j} + 3 \hat{k}$ , $b = β \hat{i} - α \hat{j} - \hat{k}$ and $c = \hat{i} - 2 \hat{j} - \hat{k}$ such that $a \cdot b = 1$ and $b \cdot c = - 3$ , then $\frac{1}{3} ((a \times b) \cdot c)$ is equal to

Let $a = \hat{i} - α \hat{j} + β \hat{k}, b = 3 \hat{i} + β \hat{j} - α \hat{k}$ and $c = - α \hat{i} - 2 \hat{j} + \hat{k}$ , where $α$ and $β$ are integers. If $a \cdot b = - 1$ and $b \cdot c = 10$ , then $(a \times b) \cdot c$ is equal to

If $a, b, c$ are vectors such that $[a b c] = 4$ then $[a \times b b \times c c \times a] =$

Let $a, b, c$ be three non-coplanar vectors and $p, q, r$ are vectors defined by the relations $p = \frac{b \times c}{[ a b c ]}, q = \frac{c \times a}{[ a b c ]}, r = \frac{a \times b}{[ a b c ]}$ then the value of the expression $(a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r$ is equal to

Let $a = \hat{i} + \hat{j} + \hat{k}$ , $b = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $c = x \hat{i} + 2 \hat{j} + 3 \hat{k}, x \in R$ . If $d$ is the unit vector in the direction of $b + c$ such that $a \cdot d = 1$ , then $(a \times b) \cdot c$ is equal to

Let a vector $a$ be coplanar with vectors $b = 2 \hat{i} + \hat{j} + \hat{k}$ and $c = \hat{i} - \hat{j} + \hat{k}$ . If $a$ is perpendicular to $d = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ , and $∣ a ∣ = 10$ . Then a possible value of $[a b c] + [a b d] + [a c d]$ is equal to:

Let $a = \hat{i} - \hat{j}$ , $b = \hat{j} - \hat{k}$ , $c = \hat{k} - \hat{i}$ . If $d$ is a unit vector such that $a \cdot d = 0 = [b c d]$ , then $d$ equals