Q.The scalar product of the vector $\hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of vectors $2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\lambda\hat{i} + 2\hat{j} + 3\hat{k}$ is equal to one. Find the value of $\lambda$.

Concept: Dot Product Angle — the scalar product of two vectors equals the product of their magnitudes times the cosine of the angle between them. Here, one vector is given and the other is a unit vector, so the dot product directly gives the cosine.

Step 1: Sum the two vectors

$\vec{S} = (2\hat{i} + 4\hat{j} - 5\hat{k}) + (\lambda\hat{i} + 2\hat{j} + 3\hat{k}) = (2+\lambda)\hat{i} + 6\hat{j} - 2\hat{k}$

Step 2: Unit vector along $\vec{S}$

$|\vec{S}| = \sqrt{(2+\lambda)^2 + 36 + 4} = \sqrt{(2+\lambda)^2 + 40}$

Unit vector $\hat{S} = \frac{\vec{S}}{|\vec{S}|}$

Step 3: Given scalar product

$\vec{A} = \hat{i} + \hat{j} + \hat{k}$ …

The problem asks: given that the dot product of $\hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of two other vectors equals 1, find $\lambda$.

Let’s unpack what’s really happening here. The scalar product (dot product) of two vectors gives a number. When one of them is a unit vector, that dot product is simply the component of the first vector along the direction of that unit vector. So the statement “scalar product equals 1” means the component of $\hat{i} + \hat{j} + \hat{k}$ along the direction of the sum vector is exactly 1.

But the sum vector itself isn’t a unit vector — we have to make it one by dividing by its magnitude. That’s the crucial step students often miss.

Step 1: Find the sum vector

Let

$\vec{a} = \hat{i} + \hat{j} + \hat{k}$

$\vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k}$

$\vec{c} = \lambda\hat{i} + 2\hat{j} + 3\hat{k}$

The sum is:

$\vec{b} + \vec{c} = (2 + \lambda)\hat{i} + (4 + 2)\hat{j} + (-5 + 3)\hat{k}$

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

Step 2: The unit vector along the sum

A unit vector in the direction of any vector $\vec{v}$ is $\frac{\vec{v}}{|\vec{v}|}$. So the unit vector along $\vec{b} + \vec{c}$ is:

$\hat{u} = \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2}}$

The denominator is the magnitude:

$|\vec{b} + \vec{c}| = \sqrt{(2 + \lambda)^2 + 36 + 4} = \sqrt{(2 + \lambda)^2 + 40}$

Step 3: Set up the dot product condition

The scalar product of $\vec{a}$ with this unit vector is given to be 1:

$\vec{a} \cdot \hat{u} = 1$

Substitute:

$(\hat{i} + \hat{j} + \hat{k}) \cdot \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 40}} = 1$

Step 4: Compute the dot product in the numerator

Dot product of $\hat{i} + \hat{j} + \hat{k}$ with $(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}$:

• $\hat{i} \cdot \hat{i}$ term: $1 \cdot (2 + \lambda) = 2 + \lambda$
• $\hat{j} \cdot \hat{j}$ term: $1 \cdot 6 = 6$
• $\hat{k} \cdot \hat{k}$ term: $1 \cdot (-2) = -2$

Sum: $(2 + \lambda) + 6 + (-2) = \lambda + 6$

So the equation becomes:

$\frac{\lambda + 6}{\sqrt{(2 + \lambda)^2 + 40}} = 1$

Step 5: Solve for $\lambda$

Multiply both sides by the denominator:

$\lambda + 6 = \sqrt{(2 + \lambda)^2 + 40}$

Square both sides (but check later for extraneous solutions): …

Method: A Dot-Product-with-a-Unit-Vector Condition

Use this when the scalar product of a vector with the unit vector along another (variable) vector is given.

Steps

Step 1: Form the sum vector

Add the two vectors whose sum defines the direction, keeping the unknown (e.g. $\lambda$) in the components.

Step 2: Build the dot-with-unit-vector equation

The unit vector along $\vec S$ is $\vec S/|\vec S|$, so 'scalar product $=1$' becomes

$\frac{\vec A\cdot\vec S}{|\vec S|}=1\quad\Longrightarrow\quad \vec A\cdot\vec S=|\vec S|.$ …

Common Mistakes

Mistake 1: Forgetting to divide by $|\vec S|$

Why it's wrong: the condition involves the unit vector along the sum, so you must normalise; using $\vec A\cdot\vec S=1$ directly gives a wrong equation and wrong $\lambda$. Correct approach: set $\dfrac{\vec A\cdot\vec S}{|\vec S|}=1$.

Mistake 2: Squaring without checking for an extraneous root …