The direction ratios of the straight line joining the points (2,3,5) and 4 6 8 are

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Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (–1, 1, 2) and (–5, –5, –2).

Let A(3, 5, – 4), B(–1, 1, 2), C(–5, –5, –2) be the vertices of ΔABC.

Direction ratios of AB are – 1 – 3, 1 – 5, 2 + 4 i.e. – 4, – 4, 6
          Dividing each by 

Direction ratios of BC are – 5 + 1, –5 –1, –2 –2 i.e. – 4, –6, –4.

  Dividing each by 

Direction Ratios of a Line

Any 3 numbers which are proportional to the direction cosines of a line are known as direction ratios of that line.

Direction ratios are denoted by (a,b and c).

Consider a vector OP-->=a î +bĵ +ck̂   (Where a, b and c are direction ratios. )

=r (cos α î + cosβ ĵ + cos γ k̂) where r=√ (a2+b2+c2)

cosα,cosβ and cosγ are direction cosines.

r cosα=a,r cosβ=b, r cosγ=c are direction ratios.

Relation between Direction Ratios and Direction Cosines

Consider a line whose one point is at originand another is at P. Considering a vector OP-->=a î +bĵ +ck̂ (equation(1))

=r cos αî+r cosβĵ+r cos γk̂

where r=√ (a2+b2+c2) (equation(2))

Comparing equations (1) and (2) :-

a=(rcos α) ,b =(r cosβ) and c=(rcosγ)(equation(3))

cos α =(a/r), cosβ=(b/r) and cosγ=(c/r) (equation(4))

Squaring and adding equation(4) :-

(a2/r2) + (b2/r2)+(c2/r2) =1

using (cos2 α + cos2 β + cos2 γ) =1

=>(a2+ b2+c2) = r2 (equation(5))

From equation(4) cos α =(a/r)= (±) (a)/(√( a2+ b2+c2)

using equation(5)

Similarly cos β=(b/r) =(±)(b)/ (√( a2+ b2+c2) and

cos γ=(c/r) =(±)(c)/ (√( a2+ b2+c2)  

usingequation (5)

Therefore cos α, cos β and cosγare the direction cosines which are written in the form of a, b and c which are direction ratios.

Relationship between direction ratios and cosines when a line passes through 2 points

One and only one line passes through two given points.

Consider a line PQ and the coordinates of points P and Q are given as P(x1, y1, z1) and Q(x2,y2,z2).

Let l,m and n are the direction cosines of the line PQ and which makes angleα,βand γ with the x,y and z-axis respectively.

If we draw perpendiculars from the point P and Q as shown in the figure (b), to XY-plane to meet at R and S. Also drawing a perpendicular from P to QS to meet at N.

Therefore cosα= (x2-x1)/(PQ)

Similarly cosβ =(y2-y1)/(PQ) and cosγ =(NQ)/(PQ) =(z2-z1)/(PQ)

Therefore the direction cosines of the line segment joining the points P(x1, y1, z1) and Q(x2, y2, z2) will be:-

PQ=(x2-x1)/(PQ) ,(y2-y1)/(PQ) +(z2-z1)/(PQ) equation(1)

Where PQ=√( x2-x1)2+( y2-y1)2+( z2-z1)2

Note:- The direction ratios of the line segment joining point P(x1, y1, z1) and

Q(x2, y2, z2) can be considered as – (x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y1 – y2, z1 – z2)

Problem:- If a line makes an angle of 90°, 135o, 45o with the positive direction ofx, y, z-axes, respectively, then find its direction cosines.

Answer:- Let l, m and be direction cosines of the line. Then l= cos90° = 0, m=cos 135o =-(1/√2) and n=cos 45o= (1/√2)

Therefore Direction cosines: [0,-(1/√2),(1/√2)]

Problem:- Find the direction cosines of a line which makes equal angles with the coordinate axes.

Answer:- Let the direction cosines of the line make an angle α with each of the coordinate axes.

l = cos α, m = cos α, n = cos α

l2+m2+n2 =1

=>cos2 α + cos2 β + cos2 γ) =1

=>3cos2 α =1

=>cos2 α = (1/3)

=>cosα = (±) (1/√3)

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes,are(±)(1/√3),(±) (1/√3),and (±) (1/√3).

Problem:- If a line has the direction ratios −18, 12, −4, then what are its direction cosines?

Answer:- If a line has direction ratios of −18, 12, and −4, then its direction cosines are

= (-18)/ (√ ((18)2+ (12)2+ (-4)2),

= (12)/ ((√ ((18)2+ (12)2+ (-4)2),

= (-4)/(√ ((18)2+ (12)2+ (-4)2)

= i.e. (-18)/ (22), (12)/ (22) and (-4)/ (22)

= (-9/11), (6/11) and (-2)/ (11)

Thus, the direction cosines are (-9/11), (6/11) and (-2)/ (11).

Problem:- Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.

Answer:- The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7).

It is known that the direction ratios of line joining the points, (x1, y1, z1) and (x2, y2, z2),are given by, (x2 − x1), (y2 − y1), and (z2 − z1).

The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., (−3, −5, and −3).

The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., (6, 10, and 6).

It can be seen that the direction ratios of BC are −2 times that of AB i.e., they areproportional.

Therefore, AB is parallel to BC. Since point B is common to AB and BC, points A, B, and C are collinear.

Problem:- Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (−1, 1, 2) and (− 5, − 5, − 2)?

Answer:- The vertices of ∆ABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).

The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., (−4), (−4), and(6).

Then,√ ((-4)2+ (-4)2+ (6)2) =√ (16+16+32)

=√ (68)

=2√ (17)

Therefore, the direction cosines of AB are

= (-4)/ (√ (-4)2 + (-4)2 + (6)2),

= (-4)/ (√ (-4)2 + (-4)2 + (6)2),

= (6)/ ((√ (-4)2 + (-4)2 + (6)2)

= (-4)/ (2√ (17), - (4)/2√ (17), (-6)/2√ (17)

= (-2)/√ (17), (-2)/√ (17), (3)/√ (17)

The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., (−4), (−6), and (−4). Therefore, the direction cosines of BC are

= (-4)/ (√ (-4)2 + (-4)2 + (6)2), (-4)/ (√ (-4)2 + (-4)2 + (6)2), (6)/ ((√ (-4)2 + (-4)2 + (6)2)

= (-4)/ (2√ (17), (-4)/(2√ (17)), (-6)/(2√ (17))

The direction ratios of CA are (−5 − 3), (−5 − 5), and (−2 − (−4)) i.e., (−8), (−10), and (2).

Therefore, the direction cosines of AC are

= (-8)/√ ((-8)2+ (10)2+ (2)2),

= (-5)/ ((-8)2+ (10)2+ (2)2),

= (2)/ ((-8)2+ (10)2+ (2)2)

=(-8)/ (2√ (42)), (-10)/ (2√ (42)), (2)/2√42))