Question

State whether the following statements are true or false. Justify your answer with a brief explanation or with a counter-example.

(a) The curve x4 + y4 = 4a2xy is symmetric with respect to the origin.

(b) The direction cosines of the line x = — y, z=0 are 1/√2, -1/√2,0

(c) The section of 2x² + y²= 2 (1 — z²) by the plane x + 2 = 0 is a hyperbola.-

(d) The xy-plane intersects the sphere

x²+ y² + z² + 2x + 2y — z = 2 in a great circle.

(e) If the projection of a line segment AB on another line is the line segment CD, then

IABI = ICDI•

Answer 1

(a) If a point $M(t,s)$ belong to the curve $x^4 + y^4 = 4a^2xy$, then $t^4 + s^4 = 4a^2ts$. Then for the point $M'(-t,-s)$ we have $(-t)^4 + (-s)^4 =t^4+s^4= 4a^2ts=4a^2(-t)(-s)$, and thus the point $M'(-t,-s)$ is also belong to the curve $x^4 + y^4 = 4a^2xy$. Consequently, the curve is symmetric with respect to the origin.

Answer: true

(b) Since the vector $v=(1,-1,0)$ is parallel to the line $x = - y,\ z=0$ and $|v|=\sqrt{2}$, we conclude that $\cos \alpha =\frac{1}{\sqrt{2}}, \ \cos \beta =-\frac{1}{\sqrt{2}}, \ \cos \gamma =\frac{0}{\sqrt{2}}=0.$

Answer: true

(c) The general equation of a hyperbola is

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

Let us find the section of $2x^2+y^2=2(1-z^2)$ by the plane $x+2=0$:

$2(-2)^2+y^2=2(1-z^2)$

$8+y^2=2-2z^2$

$y^2+2z^2=-6$

Since the eqution has no real solution, the plane $x+2=0$ does not intersect $2x^2+y^2=2(1-z^2)$. The equation $y^2+2z^2=-6$ is not a hyperbola equation.

Answer: false

(d) The xy-plane intersects the sphere $x^2+y^2+z^2+2x+2y-z=2$ in a great circle if and only if the center of this sphere belong to $xy$-plane. Let us rewrite the equation of the sphere in the the following form: $(x+1)^2+(y+1)^2+(z-\frac{1}{2})^2=2+1+1+\frac{1}{4}=\frac{17}{4}.$ It follows that $M(-1,-1,\frac{1}{2})$ is the center of the sphere. Taking into account that the third coordinate of $M$ is not equal to 0, we conclude that the center of the sphere does not belong to the $xy$-plane, and therefore, the sphere $x^2+y^2+z^2+2x+2y-z=2$ does not intersect the $xy$-plane in a great circle.

Answer: false

(e) Let consider any line segment $AB$ with $|AB|=2.$ The projection of a line segment $AB$ on another line is the line segment $CD$ formed by the projections of the end points of the line segment $AB$ on this line. If we choose such a line $l$ that the segment $AB$ is perpendicular to $l$, then the points A and B projects on the same point $C=D$ of the line $l$, and therefore, $|CD|=0$.

Answer: false