Q.If $\vec{a}=5\hat{i}-\hat{j}-3\hat{k}$ and $\vec{b}=\hat{i}+3\hat{j}-5\hat{k}$, then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.

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. $(\vec a+2\vec b)\perp\vec a$:

3. Substitute $\vec a\cdot\vec b=-|\vec b|^2$ from step 1:

4. Take square roots (magnitudes are non-negative):

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:

Since $\vec{p}\cdot\vec{q} = \vec{q}\cdot\vec{p}$, the middle terms cancel:

5. Solve for the relation

This is exactly option (C).

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.

(D) 2x + 3y + 4z = 0-> normal (2,3,4): 8 + 15 + 24 = 47 (not 0).

Only plane (C) has a normal perpendicular to the line's direction, so the line is parallel to it (in fact, since the point (3,4,5) satisfies x - 2y + z = 3 - 8 + 5 = 0, the line actually lies in that plane - which is the limiting case of being parallel; it is nevertheless the only option satisfying the parallelism condition).

ANSWER: C

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.

which reproduces the same condition and, unlike the slope derivation, stays valid even when a line is vertical ($m=0$). $\checkmark$

Eliminating the distractors: $lm'=ml'$ (option B) is the condition for the lines to be parallel (equal slopes), not perpendicular. Options (C) and (D) mix indices across the two lines and have no geometric meaning.

ANSWER: A

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:

Since $\cos^2 \theta + \sin^2 \theta = 1$, this simplifies to:

Therefore, the equation is:

Step 1 — Translate the given condition

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

Step 2 — Compute $\vec a\cdot\vec b$

With $\vec a = 2\hat i+2\hat j-\hat k$ and $\vec b=\alpha\hat i+\beta\hat j+2\hat k$:

Step 3 — Solve for $\alpha+\beta$

Setting this to zero:

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 3 — Solve

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

The official key designates option (B) as correct for this item; solving the perpendicularity condition directly gives $a=4$ or $a=-1$ (note: substituting $a=-2$ into $\vec{BA}\cdot\vec{BC}$ gives $6\neq 0$, so it does not satisfy perpendicularity — this is flagged for a follow-up key check).