Q.Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$.

Concept and Intuition

When we talk about the projection of a vector onto an axis, we mean: If you shine a light straight down onto that axis, how long is the shadow?

For a coordinate axis (like the z‑axis), this is just the component of the vector along that axis. The z‑axis is the direction of the unit vector $\hat{k}$. So the projection of any vector $\vec{v} = a\hat{\imath} + b\hat{\jmath} + c\hat{k}$ onto the z‑axis is simply the scalar $c$ (the coefficient of $\hat{k}$).

The magnitude of that projection is $|c|$ — because magnitude is always non‑negative.

Step‑by‑Step Reasoning

1. Identify the vector We are given

This means the components are:

• along $x$: $-1$
• along $y$: $2$
• along $z$: $-1$

2. Projection onto the z‑axis The z‑axis is the direction of $\hat{k}$. The projection of $\vec{v}$ onto $\hat{k}$ is the dot product $\vec{v} \cdot \hat{k}$, which picks out the $z$‑component:

So the projection (a signed scalar) is $-1$.

3. Magnitude of the projection Magnitude means absolute value (length, never negative):

4. Match with options The options are: (A) 2 (B) $\frac{1}{\sqrt{6}}$ (C) 1 (D) $-\frac{1}{\sqrt{6}}$ Clearly, $1$ corresponds to option (C).

Concept & Intuition

The projection of one vector onto another tells us how much of the first vector lies in the direction of the second. It’s like measuring the length of the shadow of $\vec{a}$ when the light shines perpendicularly onto $\vec{b}$. The formula uses the dot product because the dot product captures the component of $\vec{a}$ along $\vec{b}$, and we divide by the length of $\vec{b}$ to scale that component into actual units of length.

Step-by-step solution

1. Recall the projection formula The scalar projection of $\vec{a}$ onto $\vec{b}$ is

This gives the length of the projection (with sign indicating direction).

2. Compute the dot product Given

the dot product is

3. Find the magnitude of $\vec{b}$

4. Set the projection equal to 4

Multiply both sides by 7:

Subtract 16:

5. Verify with the options $\lambda = 4$ corresponds to option (A). But wait — let’s double-check: If $\lambda = 4$, then $\vec{a} \cdot \vec{b} = 16 + 12 = 28$, and $28/7 = 4$. Correct.

Concept & Intuition

We have a line $L_1$ through $A$ with direction $\vec{b}$. To find the perpendicular distance from an external point $P$ to this line, imagine dropping a perpendicular from $P$ onto $L_1$. The distance is the length of the component of $\overrightarrow{AP}$ that is perpendicular to $\vec{b}$. The cross product $\overrightarrow{AP} \times \vec{b}$ gives a vector whose magnitude equals the area of the parallelogram spanned by $\overrightarrow{AP}$ and $\vec{b}$. Dividing by $|\vec{b}|$ gives the height of that parallelogram — exactly the perpendicular distance we need.

Step-by-step solution

1. Find the vector from point A on the line to point P

   $\overrightarrow{AP} = \vec{p} - \vec{a} = (\hat{i}+2\hat{j}+3\hat{k}) - (4\hat{i}+2\hat{j}+2\hat{k})$

   $= (1-4)\hat{i} + (2-2)\hat{j} + (3-2)\hat{k} = -3\hat{i} + 0\hat{j} + 1\hat{k}$.

2. Recall the formula for perpendicular distance

   The distance $d$ from point $P$ to line through $A$ with direction $\vec{b}$ is

3. Compute the cross product $\overrightarrow{AP} \times \vec{b}$ $\overrightarrow{AP} = (-3, 0, 1)$, $\vec{b} = (2, 3, 6)$. Using the determinant:

$= \hat{i}(0 - 3) - \hat{j}(-18 - 2) + \hat{k}(-9 - 0)$

$= -3\hat{i} - (-20)\hat{j} - 9\hat{k} = -3\hat{i} + 20\hat{j} - 9\hat{k}$.

4. Find the magnitude of the cross product

   $|\overrightarrow{AP} \times \vec{b}| = \sqrt{(-3)^2 + 20^2 + (-9)^2} = \sqrt{9 + 400 + 81} = \sqrt{490}$.

5. Find the magnitude of $\vec{b}$

   $|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

6. Compute the distance

The line joins $R(0,-1,3)$ and $Q(2,-3,-1)$, so a direction vector is

A general point on the line is

The vector from $P(1,8,4)$ to this point is

At the foot of the perpendicular, $\vec{PL}\cdot\vec{d} = 0$:

Substitute back: