Lf A-2-1-9- B-4-1- -3- C-0-7-6 and in the - ABC the equation of the medlan through C is xa-y-7b-z-6c then a-b-c-

The correct option is B 10 'D' be midpoint of AB= 4+2 2 , 1+1 2 , 9 3 2 =( 1,1,3) Equation of CD: x 0 0+1 = y 1 7 1 = z 3 ...

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Standard XII

Mathematics

Theorems for Differentiability

lf A=2,1,9,...

Question

lf $A = (2, 1, 9), B = (- 4, 1, - 3), C = (0, 7, 6)$ and in the $Δ A B C$ the equation of the medlan through $C$ is $\frac{x}{a} = \frac{y - 7}{b} = \frac{z - 6}{c}$ then $a + b + c =$

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Solution

The correct option is B $10$
$^{'} D^{'}$ be midpoint of $A B = \frac{- 4 + 2}{2}, \frac{1 + 1}{2}, \frac{9 - 3}{2}$
$= (- 1, 1, 3)$
Equation of $C D : \frac{x - 0}{0 + 1} = \frac{y - 1}{7 - 1} = \frac{z - 3}{6 - 3}$
$\frac{x}{1} = \frac{y - 1}{6} = \frac{z - 3}{3}$
$\Rightarrow a = 1, b = 6, c = 3$
$a + b + c = 1 + 6 + 3 = 10$

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Similar questions

Q. If

A (2, 1, 9), B (- 4, 1, - 3), C (0, 7, 6)

are the vertices of

△ A B C

and the equation of the median through

C

\frac{x}{a} = \frac{y - 7}{b} = \frac{z - 6}{- 3},

then

| a + b |

is equal to

Q. HCF of

x^{6} y^{4} z^{2}

x^{6} y^{4} z^{6}

and

x^{7} y^{9} z^{4}

x^{a} y^{b} z^{c}

. Then

a - b - c

= .

Q. Add (in vertical arrangement):

(1) xy + yz + zx, 9zx + 7yz + 3yx
(2) 2x + 3y, 6x – 2y, –4x + 12y – z
(3) a²b + b²c + c²a, 10ac² + 2ba² – 16cb²
(4) 15mn – 6ab + 7abc, abc – 8nm + 20ba

Simplify:

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

If
A = {1, 2, 3, 4}

B = {3, 4, 5, 6, 7}

C = {6, 7, 8, 9}

D = {7, 8, 9, 10}

$\begin{matrix} C o l u m n A & C o l u m n B 1. A \cup B & a . {3, 4, 6, 7} 2. A \cap B & b . \emptyset 3. (B \cap C) \cup D & c . {3, 4} 4. B \cup C \cup D & d . {3, 4, 5, 6, 7, 8, 9, 10} 5. (A \cap B) \cap (B \cup C) & e . {1, 2, 3, 4, 5, 6, 7} f . {6, 7, 8, 9, 10} g . {7} \end{matrix}$

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lf A=(2,1,9),B=(−4,1,−3),C=(0,7,6) and in the ΔABC the equation of the medlan through C is xa=y−7b=z−6c then a+b+c=

The correct option is B 10′D′ be midpoint of AB=−4+22,1+12,9−32=(−1,1,3)Equation of CD:x−00+1=y−17−1=z−36−3x1=y−16=z−33⇒a=1,b=6,c=3a+b+c=1+6+3=10

lf $A = (2, 1, 9), B = (- 4, 1, - 3), C = (0, 7, 6)$ and in the $Δ A B C$ the equation of the medlan through $C$ is $\frac{x}{a} = \frac{y - 7}{b} = \frac{z - 6}{c}$ then $a + b + c =$