25.05.2018 Views

vector

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

Vectors 385<br />

Normal to (2), 2<br />

= (z – x 2 – y 2 + 3)<br />

=<br />

<br />

<br />

<br />

ˆ <br />

<br />

x y z<br />

ˆ ˆ 2 2<br />

i j k ( z– x – y 3)<br />

= – 2 x î – 2 y ĵ + ˆk<br />

Normal to (2) at (2, – 1, 2), 2<br />

= – 4î + 2 ĵ + ˆk ...(4)<br />

1.<br />

2 = | 1|| 2|cos<br />

<br />

1.<br />

2<br />

(4 iˆ –2ˆj 4 kˆ).(– 4iˆ 2 ˆj kˆ) –16 –44<br />

cos = =<br />

=<br />

| 1|| 2|<br />

|4 iˆ –2ˆj 4 kˆ||– 4iˆ 2 ˆj kˆ|<br />

16 416 16 41<br />

= –16 –8<br />

=<br />

6 21 3 21<br />

–1 –8 <br />

= cos <br />

3 21 <br />

–1 –8 <br />

Hence the angle between (1) and (2) cos <br />

Ans<br />

3 21 <br />

EXERCISE 5.6<br />

2<br />

3<br />

t<br />

t<br />

1. The coordinates of a moving particle are given by x = 4t and y = 3 6 t . Find the<br />

2<br />

6<br />

velocity and acceleration of the particle when t = 2 secs. Ans. 4.47, 2.24<br />

2. A particle moves along the curve<br />

x = 2t 2 , y = t 2 – 4t and z = 3t – 5<br />

where t is the time. Find the components of its velocity and acceleration at time t = 1, in the<br />

<br />

8 14 14<br />

direction i 3 j 2 k.<br />

(Nagpur, Summer 2001) Ans. , <br />

7 7<br />

3. Find the unit tangent and unit normal <strong>vector</strong> at t = 2 on the curve x = t 2 – 1, y = 4t – 3,<br />

z = 2t 2 – 6t where t is any variable.<br />

Ans.<br />

1 1 <br />

(2 i 2 j k), (2i 2 k)<br />

3 3 5<br />

<br />

d<br />

<br />

dG d F <br />

4. Prove that ( F G)<br />

F G<br />

dt dt dt<br />

<br />

5. Find the angle between the tangents to the curve 2 3<br />

r t i 2 t j t k,<br />

at the points t = ± 1.<br />

Ans. 1 9<br />

cos<br />

<br />

<br />

<br />

17<br />

<br />

6. If the surface 5x 2 – 2byz = 9x be orthogonal to the surface 4x 2 y + z 3 = 4 at the point (1, –1, 2)<br />

then b is equal to<br />

(a) 0 (b) 1 (c) 2 (d) 3 (AMIETE, Dec. 2009) Ans. (b)<br />

5.25 SCALAR AND VECTOR POINT FUNCTIONS<br />

Point function. A variable quantity whose value at any point<br />

N<br />

^<br />

in a region of space depends upon the position of the point, is<br />

R<br />

Q<br />

called a point function. There are two types of point functions.<br />

n r<br />

(i) Scalar point function. If to each point P (x, y, z) of a<br />

region R in space there corresponds a unique scalar f (P), then f is<br />

P<br />

called a scalar point function. For example, the temperature<br />

distribution in a heated body, density of a body and potential due to gravity are the examples of<br />

a scalar point function.<br />

(ii) Vector point function. If to each point P (x, y, z) of a region R in space there corresponds<br />

a unique <strong>vector</strong> f (P), then f is called a <strong>vector</strong> point function. The velocity of a moving fluid,<br />

gravitational force are the examples of <strong>vector</strong> point function.<br />

(U.P., I Semester, Winter 2000)<br />

= c<br />

<br />

+ d = c

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!