Q.If the vertices A, B, C of a triangle ABC are $(1, 2, 3), (-1, 0, 0), (0, 1, 2),$ respectively, then find $\angle ABC. [\angle ABC$ is the angle between the vectors $\vec{BA}$ and $\vec{BC}].$

$\mathbf a=\hat i+2\hat j+2\hat k,\ \mathbf b=\hat i+2\hat j-2\hat k$.

$\mathbf a\cdot\mathbf b=1\cdot1+2\cdot2+2\cdot(-2)=1+4-4=1.$

$|\mathbf a|=\sqrt{1+4+4}=3,\quad |\mathbf b|=\sqrt{1+4+4}=3.$ …

The key idea: The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ satisfies $\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$. This works because the dot product measures how much the vectors point in the same direction, scaled by their lengths. Here, we just compute the dot product and magnitudes carefully.

1. Write the vectors in component form.

   $\vec{a} = \hat{\imath} - \hat{\jmath}$ means components $(1, -1, 0)$.

   $\vec{b} = \hat{\jmath} - \hat{k}$ means components $(0, 1, -1)$.

2. Compute the dot product.

   $\vec{a}\cdot\vec{b} = (1)(0) + (-1)(1) + (0)(-1) = 0 - 1 + 0 = -1$.

3. Compute the magnitudes.

   $|\vec{a}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1+1} = \sqrt{2}$.

   $|\vec{b}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0+1+1} = \sqrt{2}$.

4. Find $\cos\theta$.

   $\cos\theta = \frac{-1}{\sqrt{2}\cdot\sqrt{2}} = \frac{-1}{2}$.

5. Determine the angle. …

Concept: cos(theta) = (a . b) / (|a| |b|)

a = 2i + 2j - k , b = 6i - 3j + 2k

Dot product:

a . b = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4

Magnitudes:

|a| = sqrt(4 + 4 + 1) = sqrt(9) = 3

|b| = sqrt(36 + 9 + 4) = sqrt(49) = 7 …

The key idea: direction cosines are the components of a unit vector along the line. So the given triples are already unit vectors. The angle between the lines is simply the angle between these two unit vectors, found via their dot product.

A common pitfall: the dot product formula gives $\cos\theta$, and $\theta$ is taken as the acute angle between lines (since lines have no direction, the angle between them is defined as the smaller angle, between $0$ and $\frac{\pi}{2}$). If $\cos\theta$ turns out negative, the acute angle is $\pi - \theta$.

Let the two direction vectors be:

1. Verify they are unit vectors For $\mathbf{a}$:

For $\mathbf{b}$:

Both are indeed unit vectors, so the given triples are valid direction cosines.

2. Compute the dot product

3. Find the angle …

The key idea is to express both the dot product and the magnitude of the cross product in terms of the magnitudes of the vectors and the angle between them. The dot product is $|\vec{a}||\vec{b}|\cos\theta$, and the magnitude of the cross product is $|\vec{a}||\vec{b}|\sin\theta$. Setting them equal gives a simple trigonometric equation.

1. Write the given condition using standard formulas. For any two vectors $\vec{a}$ and $\vec{b}$, with angle $\theta$ between them ($0 \le \theta \le \pi$), we have:

and

The problem states:

2. Simplify the equation. Assuming neither vector is zero (if either were zero, the angle is undefined, but the condition would hold trivially; however, the problem implies non-zero vectors), we can divide both sides by $|\vec{a}|\,|\vec{b}|$ (which is positive), obtaining:

3. Solve the trigonometric equation. The equation $\cos\theta = \sin\theta$ can be rewritten as $\tan\theta = 1$ (provided $\cos\theta \neq 0$; if $\cos\theta = 0$, then $\sin\theta = 1$ or $-1$, which doesn't satisfy equality). The principal solution in the range $0 \le \theta \le \pi$ is: $\theta = \frac{\pi}{4}.$ …

Step 1 — Extract the vectors.

The line $\dfrac{x-5}{2}=\dfrac{y+1}{-1}=\dfrac{z+4}{1}$ has direction vector

The plane $3x-4y-z+5=0$ has normal

Step 2 — Use the correct formula.

The angle $\theta$ between a line and a plane is measured from the plane, so it is the complement of the angle between $\vec b$ and $\vec n$:

Step 3 — Compute.

Step 4 — Convert to the $\cos^{-1}$ form used in the options. …

The key here is to recall how two vectors add. When you have two vectors of equal magnitude, their resultant’s magnitude depends only on the angle between them. The phrase “square of resultant” means the square of the magnitude of the sum, and “product” means the product of the magnitudes of the two vectors. Since the vectors are equal, that product is just the square of one vector’s magnitude. This is a classic setup for using the parallelogram law of vector addition.

Let’s work through it step by step.

1. Set up the notation.

   Let each electric field vector have magnitude $E$. So $|\vec{E}_1| = |\vec{E}_2| = E$.

   The resultant vector is $\vec{R} = \vec{E}_1 + \vec{E}_2$.

   The square of the resultant is $|\vec{R}|^2$, and the product of the two vectors’ magnitudes is $E \cdot E = E^2$.

2. Translate the given condition into an equation.

   The problem says: “The square of resultant is three times their product.”

   So:

3. Use the formula for the magnitude of the sum of two vectors. For any two vectors with magnitudes $a$ and $b$ and angle $\theta$ between them:

Here $a = b = E$, so:

4. Set this equal to the given value.

Since $E^2 \neq 0$, divide both sides by $E^2$:

5. Solve for $\cos\theta$. …

The key idea here is that mutually perpendicular unit vectors give us two clean facts: each vector has length 1, and their dot product is zero. This turns what looks like a messy expansion into simple arithmetic.

When you see a dot product of two linear combinations of perpendicular unit vectors, you never need to guess coordinates — just expand using the distributive property and apply the three basic rules:

• $\vec{a}\cdot\vec{a} = |\vec{a}|^2 = 1$
• $\vec{b}\cdot\vec{b} = |\vec{b}|^2 = 1$
• $\vec{a}\cdot\vec{b} = 0$

Let’s work through it.

1. Expand the dot product using the distributive law (FOIL works here):

2. Pull out the scalar coefficients — dot product is bilinear, so constants factor out:

3. Apply the unit/perpendicular facts:

   • $\vec{a}\cdot\vec{a} = 1$
   • $\vec{b}\cdot\vec{b} = 1$
   • $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a} = 0$

   So the expression becomes: …

The vector $\vec Q\times\vec R$ is perpendicular to both $\vec Q$ and $\vec R$. A dot product of two perpendicular vectors is zero:

$(\vec Q\times\vec R)\cdot\vec Q=0.$ …