Q.Find the acute angle between the line $\vec{r}=(\hat{i}+2\hat{j}+2\hat{k})+\lambda(2\hat{i}+3\hat{j}-6\hat{k})$ and the plane $\vec{r}\cdot(2\hat{i}-\hat{j}+\hat{k})=0$.

The line's direction is perpendicular to the plane's normal (parallel to plane) and a point of the line satisfies the plane's equation — exactly the criterion stated in Reason.

The line $\dfrac{x-1}{3}=\dfrac{y-2}{11}=\dfrac{z+1}{11}$ has direction ratios $(3,11,11)$ and passes through $(1,2,-1)$. The plane $11x-3z-14=0$ has normal $(11,0,-3)$.

Parallel check: $\vec d\cdot\vec n = (3)(11)+(11)(0)+(11)(-3) = 33+0-33=0$, so the line is parallel to the plane.

Point-on-plane check: substitute $(1,2,-1)$: $11(1)-3(-1)-14 = 11+3-14=0$, so the point lies on the plane.

Both conditions hold, so the line lies entirely in the plane — Assertion (A) is TRUE. Reason (R) states exactly this criterion (parallel + one point on the plane $\Rightarrow$ line lies in plane), which is a true general fact and is precisely why (A) holds — so (R) correctly explains (A).

A line is parallel to a plane when its direction vector is perpendicular to the plane's normal vector, i.e. their dot product is zero.

The line $\vec r=(\hat i-2\hat j+\hat k)+\lambda(2\hat i+\hat j+2\hat k)$ has direction vector $\vec d=(2,1,2)$.

The plane $\vec r\cdot(3\hat i-2\hat j+m\hat k)=14$ has normal vector $\vec n=(3,-2,m)$.

For the line to be parallel to the plane: $\vec d\cdot\vec n=0$.

$2(3)+1(-2)+2(m)=0$

$6-2+2m=0$

$4+2m=0\ \Rightarrow\ m=-2$

For the line to lie on the plane, its direction must be perpendicular to the plane's normal and its point must satisfy the plane equation; this gives $k=7$.

The line is $\dfrac{x-4}{1}=\dfrac{y-2}{1}=\dfrac{z-k}{2}$, with direction ratios $(1,1,2)$ and passing through the point $(4,2,k)$.

The plane is $2x-4y+z=7$, with normal vector $(2,-4,1)$.

Condition 1 (line parallel to plane): direction vector $\perp$ normal, i.e. dot product $=0$:

$(1)(2)+(1)(-4)+(2)(1) = 2-4+2 = 0$ ✓ (satisfied for all $k$, so the line is parallel to the plane).

Condition 2 (the point of the line lies in the plane): substitute $(4,2,k)$ into the plane equation:

$2(4) - 4(2) + k = 7$

$8 - 8 + k = 7$

$k = 7$

line parallel to plane ⇒ direction vector ⊥ plane's normal

Direction vector of the line: $(-K,2,1)$. Normal to the plane $\vec r\cdot(2\hat i+3\hat j-4\hat k)+7=0$: $(2,3,-4)$.

Parallelism requires $(-K)(2)+2(3)+1(-4)=0$:

$-2K+6-4=0\implies -2K+2=0\implies K=1$

line is parallel to a plane when its direction ratios are perpendicular to the plane's normal

The line has direction ratios $(3,4,5)$. Test each plane's normal $\vec n$ for $\vec n\cdot(3,4,5)=0$:

1. $(2,3,4)$: $6+12+20=38\ne0$
2. $(3,4,5)$: $9+16+25=50\ne0$
3. $(1,1,1)$: $3+4+5=12\ne0$
4. $(2,1,-2)$: $6+4-10=0$ ✓

line parallel to a plane ⇒ direction ratios ⟂ normal

The line $\dfrac{x+R}{3p}=\dfrac{y-3}{2}=\dfrac{z-4}{1}$ has direction ratios $(3p,\,2,\,1)$ — the constant $R$ in the numerator is only a shift of origin along the line and plays no role in its direction, so its (garbled/illegible) value does not affect the answer.

The plane $2x+3y-4z+7=0$ has normal vector $(2,3,-4)$.

A line is parallel to a plane iff its direction vector is perpendicular to the plane's normal, i.e. their dot product is $0$:

$3p(2)+2(3)+1(-4)=0\implies 6p+6-4=0\implies 6p=-2\implies p=-\dfrac{1}{3}$