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) true

b) true

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

c) false

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.

d) false

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: $(x+1)^ 2 +(y+1)^ 2 +(z−1/2 ​ )^ 2 =2+1+1+1/4 =17/4$ . It follows that $M(−1,−1, 1/2 ​ )$  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.

e) false

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, $|CD|=0$.