Q.Prove that $(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2$, if and only if $\vec{a},\vec{b}$ are perpendicular, given $\vec{a}\neq\vec{0},\vec{b}\neq\vec{0}$.

Concept & Intuition

Two vectors are perpendicular exactly when their dot product is zero. Expanding each condition and combining them relates the magnitudes of $\vec a$ and $\vec b$.

Step-by-step solution

1. $(\vec a+\vec b)\perp\vec b$:

2. $(\vec a+2\vec b)\perp\vec a$: …

Concept & Intuition

When a point divides a segment internally in a given ratio, its position vector is a weighted average of the endpoints. When it divides externally, the weights have opposite signs. Here, R divides PQ internally in the ratio 2:3 (meaning PR:RQ = 2:3), and S divides PQ externally in the same ratio (meaning PS:SQ = 2:3, but S lies outside the segment). The vectors $\overrightarrow{OR}$ and $\overrightarrow{OS}$ are perpendicular exactly when their dot product is zero. That dot product will involve $\vec{p}$ and $\vec{q}$, and simplifying it yields a condition on their magnitudes.

Step-by-step solution

1. Write the position vector of R (internal division) For internal division in the ratio $m:n$, the position vector is $\frac{n\vec{p} + m\vec{q}}{m+n}$. Here $m=2$, $n=3$ (since PR:RQ = 2:3, the point is closer to Q). So

2. Write the position vector of S (external division) For external division in the ratio $m:n$, the formula is $\frac{-n\vec{p} + m\vec{q}}{m-n}$ (or equivalently $\frac{n\vec{p} - m\vec{q}}{n-m}$). Using $m=2$, $n=3$:

3. Set the dot product to zero for perpendicularity

Substitute:

Multiply both sides by 5:

4. Expand the dot product Using the distributive property: $$$$

Step 1 — Slopes from the general form.

For $ax+by=c$ the slope is $-\dfrac{a}{b}$. Hence

Notice $n$ and $n'$ never appear — they only translate the lines, so they cannot affect perpendicularity.

Step 2 — Apply the perpendicularity condition.

Two non-vertical lines are perpendicular iff the product of their slopes is $-1$:

Step 3 — Clear the denominator.

Step 4 — Why this is the right form (and a vector cross-check).

The normal vectors of the two lines are $\vec{n_1}=(l,m)$ and $\vec{n_2}=(l',m')$. Two lines are perpendicular exactly when their normals are perpendicular, i.e. …

We start with the given line: $x \sec \theta + y \csc \theta = a$. To find its slope, rewrite it in the form $y = mx + c$.

Recall $\sec \theta = \frac{1}{\cos \theta}$ and $\csc \theta = \frac{1}{\sin \theta}$. So the equation becomes:

Multiply through by $\cos \theta \sin \theta$:

Now solve for $y$:

So the slope of the given line is $-\tan \theta$.

1. Slope of the perpendicular line If two lines are perpendicular, the product of their slopes is $-1$. Let the slope of the required line be $m$. Then:

So the required line has slope $\cot \theta$.

2. Equation using point-slope form The line passes through $(a \cos^3 \theta, a \sin^3 \theta)$. Using $y - y_1 = m (x - x_1)$:

Since $\cot \theta = \frac{\cos \theta}{\sin \theta}$, multiply both sides by $\sin \theta$:

3. Rearrange to standard form Bring terms together:

Factor the right-hand side: …

Concept: A line with direction ratios (a, b, c) is parallel to the plane with normal (l, m, n) iff the direction vector is perpendicular to the normal, i.e. al + bm + cn = 0.

The line (x-3)/4 = (y-4)/5 = (z-5)/6 has direction (4, 5, 6) and passes through (3, 4, 5).

Test each plane's normal:

(A) 3x + 4y + 5z = 7 -> normal (3,4,5): 12 + 20 + 30 = 62 (not 0).

(B) x + y + z = 2 -> normal (1,1,1): 4 + 5 + 6 = 15 (not 0).

(C) x - 2y + z = 0 -> normal (1,-2,1): 4 - 10 + 6 = 0 -> direction is perpendicular to the normal. YES. …

Step 1 — Translate the given condition

Squaring both sides of $|\vec a+\vec b| = |\vec a-\vec b|$:

Step 2 — Compute $\vec a\cdot\vec b$ …

Step 1 — Write the orthogonality condition

$\vec a=2\hat i+\lambda\hat j+\hat k$ and $\vec b=\hat i+2\hat j+3\hat k$ are orthogonal when $\vec a\cdot\vec b=0$.

Step 2 — Compute the dot product …

Step 1 — Form vectors from $B$

Step 2 — Apply the perpendicularity condition

Since $\angle B = 90°$, $\vec{BA}\cdot\vec{BC}=0$:

Step 3 — Expand and solve

Step 4 — Match to the given option …