Q.The value of $\hat{i}\cdot(\hat{j}\times\hat{k})+\hat{j}\cdot(\hat{i}\times\hat{k})+\hat{k}\cdot(\hat{i}\times\hat{j})$ is (A) 0 (B) -1 (C) 1 (D) 3

The problem asks us to find the value of $x$ given four vectors and a condition involving a vector triple product and perpendicularity. The most efficient way to approach this is by using the vector triple product identity, which simplifies the calculation significantly.

Concept and Intuition

1. Vector Triple Product Identity: Calculating a cross product twice, like $(\vec{a} \times \vec{b}) \times \vec{c}$, can be tedious if done directly. A powerful identity exists that converts this into a combination of dot products and scalar multiplications of vectors. This identity is:

   $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$

   This formula is extremely useful because dot products are scalar values, making the subsequent vector operations much simpler. It essentially tells us that the resulting vector lies in the plane formed by $\vec{a}$ and $\vec{b}$.

2. Perpendicularity Condition: Two non-zero vectors, say $\vec{u}$ and $\vec{v}$, are perpendicular (or orthogonal) if and only if their dot product is zero.

   $\vec{u} \cdot \vec{v} = 0$

   This is a fundamental property of the dot product, as $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$, and for perpendicular vectors, $\theta = 90^\circ$, so $\cos\theta = 0$.

Our strategy will be to first use the vector triple product identity to find the vector $(\vec{a} \times \vec{b}) \times \vec{c}$ in terms of $\vec{i}, \vec{j}, \vec{k}$. Then, we will take the dot product of this resulting vector with $\vec{d}$ and set it to zero to solve for $x$.

Step-by-step Derivation

1. Calculate the required dot products:

   We need $\vec{a} \cdot \vec{c}$ and $\vec{b} \cdot \vec{c}$ for the vector triple product identity.

   Given $\vec{a} = 2\vec{i} - 3\vec{j} + \vec{k}$ and $\vec{c} = \vec{i} - \vec{j}$.

   $\vec{a} \cdot \vec{c} = (2)(1) + (-3)(-1) + (1)(0) = 2 + 3 + 0 = 5$

   Given $\vec{b} = \vec{i} + 2\vec{j} - 3\vec{k}$ and $\vec{c} = \vec{i} - \vec{j}$.

   $\vec{b} \cdot \vec{c} = (1)(1) + (2)(-1) + (-3)(0) = 1 - 2 + 0 = -1$

2. Apply the vector triple product identity:

   Let $\vec{R} = (\vec{a} \times \vec{b}) \times \vec{c}$. Using the identity:

   $\vec{R} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$

   Substitute the dot product values and the given vectors $\vec{a}$ and $\vec{b}$:

   $\vec{R} = (5)(\vec{i} + 2\vec{j} - 3\vec{k}) - (-1)(2\vec{i} - 3\vec{j} + \vec{k})$ …

The key idea: when you know both the cross product and the dot product of an unknown vector with a given vector, you can reconstruct it using the identity $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$. This lets you isolate $\vec{b}$ without solving three simultaneous equations from scratch.

1. Set up what we know. $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{c} = \hat{j} - \hat{k}$, and we have

2. Take the cross product of $\vec{a}$ with the first equation.

The left side expands using the vector triple product identity:

Here $\vec{a} \cdot \vec{a} = 1^2 + 1^2 + 1^2 = 3$, and $\vec{a} \cdot \vec{b} = 3$ is given. So

3. Compute $\vec{a} \times \vec{c}$.

Expanding:

$\hat{i}(1 \cdot (-1) - 1 \cdot 1) - \hat{j}(1 \cdot (-1) - 1 \cdot 0) + \hat{k}(1 \cdot 1 - 1 \cdot 0)$

$= \hat{i}(-1 - 1) - \hat{j}(-1 - 0) + \hat{k}(1 - 0)$

$= -2\hat{i} + \hat{j} + \hat{k}.$

4. Plug into the equation from step 2.

Since $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, we have $3\vec{a} = 3\hat{i} + 3\hat{j} + 3\hat{k}$.

So

5. Solve for $\vec{b}$. …

We are given three vectors $\vec{a}, \vec{b}, \vec{c}$ each with the same squared length:

$\vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c} = 5$.

We also have a sum of three squared magnitudes equal to 50.

We need the sum of the pairwise dot products.

Concept & Intuition

When you see expressions like $|\vec{a} + \vec{b} - \vec{c}|^2$, think of expanding them using the dot product:

$|\vec{x}|^2 = \vec{x} \cdot \vec{x}$.

This turns geometric lengths into algebraic sums of dot products. Since we know each vector’s self-dot, the only unknowns are the cross terms $\vec{a} \cdot \vec{b}$, $\vec{b} \cdot \vec{c}$, $\vec{c} \cdot \vec{a}$. The given equation becomes a linear equation in these unknowns.

Step-by-step solution

1. Expand each squared magnitude For any vectors, $|\vec{u} + \vec{v} - \vec{w}|^2 = (\vec{u}+\vec{v}-\vec{w})\cdot(\vec{u}+\vec{v}-\vec{w})$. So:

Similarly:

2. Sum the three expansions Add the three expressions term by term.
   • The self-dot terms: each of $\vec{a}\cdot\vec{a}$, $\vec{b}\cdot\vec{b}$, $\vec{c}\cdot\vec{c}$ appears in all three expansions, so they contribute $3 \times (5+5+5) = 3 \times 15 = 45$.
   • The cross terms:
     • $2\vec{a}\cdot\vec{b}$ appears in the first expansion, $-2\vec{a}\cdot\vec{b}$ in the second, and $-2\vec{a}\cdot\vec{b}$ in the third. Sum = $2 - 2 - 2 = -2$ times $\vec{a}\cdot\vec{b}$.
     • $2\vec{b}\cdot\vec{c}$ appears in the second, $-2\vec{b}\cdot\vec{c}$ in the first, and $-2\vec{b}\cdot\vec{c}$ in the third. Sum = $2 - 2 - 2 = -2$ times $\vec{b}\cdot\vec{c}$. …

The problem gives us two vectors and an equation that looks messy at first glance. The key is to recognize the structure: $\vec{b} \times (\vec{a} \times \vec{b})$ is a vector triple product. There is a standard identity for this that will let us simplify the left-hand side dramatically, and then we can compare it with the right-hand side to extract $k$ and $l$.

Let’s recall the vector triple product identity:

Applying this with $\vec{u} = \vec{b}$, $\vec{v} = \vec{a}$, and $\vec{w} = \vec{b}$, we get:

That’s $|\vec{b}|^2 \vec{a} - (\vec{a} \cdot \vec{b})\vec{b}$.

The problem states this equals $\frac{\vec{a} - k\vec{b}}{l}$. So we have:

Now we can work step by step.

1. Compute $|\vec{b}|^2$ and $\vec{a} \cdot \vec{b}$.

   $\vec{b} = 2\vec{i} - \vec{j} - \vec{k}$, so

   $|\vec{b}|^2 = 2^2 + (-1)^2 + (-1)^2 = 4 + 1 + 1 = 6$.

   $\vec{a} = 3\vec{i} + 4\vec{j} - 5\vec{k}$, so

   $\vec{a} \cdot \vec{b} = (3)(2) + (4)(-1) + (-5)(-1) = 6 - 4 + 5 = 7$.

2. Substitute these into the left-hand side.

3. Set this equal to the right-hand side.

4. Multiply both sides by $l$ to clear the denominator.

5. Compare coefficients of $\vec{a}$ and $\vec{b}$. Since $\vec{a}$ and $\vec{b}$ are not parallel (you can check: their cross product is non-zero), the coefficients must match independently. For $\vec{a}$: $6l = 1$, so $l = \frac{1}{6}$. …

The key idea: when a problem says "$\vec{a}$ is perpendicular to the plane containing vectors $\vec{u}$ and $\vec{v}$", it means $\vec{a}$ is parallel to $\vec{u} \times \vec{v}$. The magnitude of the projection of a vector $\vec{w}$ onto $\vec{a}$ is then the same as the magnitude of its projection onto $\vec{u} \times \vec{v}$, because projection magnitude depends only on direction, not on the length of $\vec{a}$.

Let’s work through it.

1. Find a vector perpendicular to the plane. Let $\vec{u} = 2\vec{i} - \vec{j} + 3\vec{k}$ and $\vec{v} = -\vec{i} + 3\vec{j} + 2\vec{k}$. Their cross product $\vec{u} \times \vec{v}$ is perpendicular to both, hence perpendicular to the plane they span. Compute:

So any vector $\vec{a}$ perpendicular to the plane is parallel to $\vec{n} = -11\vec{i} - 7\vec{j} + 5\vec{k}$.

2. Magnitude of projection onto $\vec{a}$. The magnitude of the projection of $\vec{w} = 3\vec{i} + 2\vec{j} - \vec{k}$ onto $\vec{a}$ is:

Since $\vec{a}$ is parallel to $\vec{n}$, the direction is the same. The ratio $|\vec{w} \cdot \vec{a}| / |\vec{a}|$ equals $|\vec{w} \cdot \vec{n}| / |\vec{n}|$ because scaling $\vec{a}$ cancels out. So we can just use $\vec{n}$.

3. Compute dot product and magnitude.

So $|\vec{w} \cdot \vec{n}| = 52$.