I. The radius of the circle x^2+y^2+z^2= 49 (A) (10)/(3 √(3)) 2 x+3 y-z-5 √(14)=0 is II. The loc

step 1 Hi, we're given here this plane here. There's the plane on the axis of y and there's the interce

I. The radius of the circle x^2+y^2+z^2= 49 (A) (10)/(3...

I. The radius of the circle $x^{2}+y^{2}+z^{2}=$ 49 (A) $\frac{10}{3 \sqrt{3}}$ $2 x+3 y-z-5 \sqrt{14}=0$ is II. The locus of the foot of the perpen- (B) $\sqrt{14}$ dicular from the origin on the variable plane through the fixed point $(2,-4,6)$ is a sphere of radius III. The distance of the line $L$ whose vector (C) 3 equation is $\mathbf{r}=2 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k}+\lambda(\mathbf{i}-\mathbf{j}$ $+4 \mathbf{k}$ ) from the plane $\pi$ whose vector equation is $\mathbf{r} \cdot(\mathbf{i}+5 \mathbf{j}+\mathbf{k})=5$, is IV. The equation of the plane which meets (D) $2 \sqrt{6}$ the axes in $A, B$ and $C$, given that the centroid of the triangle $A B C$ is the point $(\alpha, \beta, \gamma)$, is $\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=k$, where $k=$

I. The radius of the circle $x^{2}+y^{2}+z^{2}=$ 49
(A) $\frac{10}{3 \sqrt{3}}$
$2 x+3 y-z-5 \sqrt{14}=0$ is
II. The locus of the foot of the perpen-
(B) $\sqrt{14}$
dicular from the origin on the variable plane through the fixed point $(2,-4,6)$ is a sphere of radius
III. The distance of the line $L$ whose vector (C) 3 equation is $\mathbf{r}=2 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k}+\lambda(\mathbf{i}-\mathbf{j}$
$+4 \mathbf{k}$ ) from the plane $\pi$ whose vector equation is $\mathbf{r} \cdot(\mathbf{i}+5 \mathbf{j}+\mathbf{k})=5$, is
IV. The equation of the plane which meets
(D) $2 \sqrt{6}$
the axes in $A, B$ and $C$, given that the centroid of the triangle $A B C$ is the point $(\alpha, \beta, \gamma)$, is $\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=k$, where
$k=$

Key Concepts

Distance from a Line to a Plane

The distance from a line to a plane is defined as the shortest distance between any point on the line and the plane. In cases where the line is parallel to the plane or under certain conditions, this distance remains constant and is computed using the point-plane distance formula, considering the line’s direction and a point on the line.

Intersection of a Sphere and a Plane

When a plane intersects a sphere, their intersection is a circle. The radius of this circle can be found by computing the distance from the center of the sphere to the plane and then using the Pythagorean theorem to relate the sphere’s radius to the radius of the circle of intersection.

Foot of a Perpendicular and Locus Determination

The foot of the perpendicular from a fixed point to a plane is the point on the plane that is closest to that fixed point. When considering how this foot varies as the plane shifts under a given condition (such as passing through a fixed point), the set of all such points forms a locus, which in symmetric circumstances can take the form of a sphere.

Intercept Form of the Equation of a Plane and the Triangle Centroid

A plane that intercepts the coordinate axes can be expressed in intercept form, making the points of intersection with the axes explicit. The triangle formed by these intercepts has a centroid given by the average of the vertices' coordinates. Relating the intercepts to the centroid coordinates is a useful technique in deriving or characterizing the plane's equation.

Transcript

00:01 Hi, we're given here this plane here.

00:03 There's the plane on the axis of y and there's the intercepts a, b, and c.

00:07 There's a centroid on the triangle here, so formed, centroid coordinates are 1r -r -squith, and the plane passes to a point 4 negative 815, to find the value of r.

00:19 So for this we can say the triangle vertices here, this vertex is given as a, 0, 0.

00:27 This vertex is 0, b, 0, and this vertex is 0 .0.

00:32 0, 0.

00:34 This is going off.

00:36 So, centroid is given as we have x1 plus x2 plus x3 over 3, y1 plus y2 plus y3 over 3 and so on.

00:43 So g coordinates are given as a over 3, then b over 3, then c over 3.

00:49 Now it's given as g and g coordinates are 1 are r squared.

00:52 So we get from there a equals 3, then we have b equals 3 r, and c equals 3 r squared.

01:02 So the equation of clean is coming out to be we will have x over 3 plus y over 3 r into that form plus z over 3r that is equal to 1 and you can simplify this and we will get r square x plus r y plus z equals 3 r square now it is passing to the point point is 4 negative 8 15 so therefore uh we have put here it will be 4 r square then negative 8 r positive 15 that equals 3 r square so we can just solve this now this equation so we get r square negative 8 r positive 15 is equal to 0 or we get r r is going out to be 3 or 5 so let's check the answers for that so it is a corresponds to q and s let's do the second part now next to the line is given as parallel to the plane, this plane here, we find a value of r.

02:13 We have a direction ratio to normal.

02:16 From this, the direction ratio to normal, that is given as a equal to 1, b equal 1, and c equals 1.

02:22 This line is parallel to the plane.

02:24 So therefore, we have this ration ratio to the line parallel to the line.

02:27 So i'll just show in the figure here.

02:29 Let's say this is a plane here we have.

02:34 That is the plane.

02:35 So this is the plane here.

02:38 So it is given as the equation is x plus y plus d equal to 3.

02:43 So the normal to this, this is the normal, having reduction ratios, 1, 1, 1.

02:50 The line is parallel to the plane.

02:51 This is the line here, which is parallel to the plane.

02:54 And the reduction ratios are of the line which is parallel to this line.

02:57 It's given as r, r square, negative 12.

03:01 So we will have this vector is perpendicular to this vector here.

03:06 The dot product is 0.

03:07 So we get r, positive r square, negative 12, is equal to.

03:15 From here we get r equal to 3 or negative 4.

03:19 Check the options here.

03:21 R is 3 or negative 4.

03:24 Therefore, b corresponds to r and s.

03:31 Next, moving on to cpart.

03:37 It is given that in c part, plane is perpendicular to the line, x over 1, y over r, 3 over r squared, pass us to the region, and passes to the point...

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