The vertices B and C of a triangle ABC lie on the line, x+23=y−10=z4 such that BC=5 units. Then the area (in sq. units) of this triangle, given that the point A(1,-1,2),is:

Let a point D on BC= (3λ,−2,1,4λ) AD→=(3λ−3)i∧+2j∧+(4λ−2)k∧∵ AD→ ⊥ BC→ , ∴AD→ . BC→=0⇒(3λ−3)+3+2(0)+(4λ−2)4=0⇒λ=1725Hence, D=(125,1,6825)|AD→|=(125−1)2+(2)2+(6825−2)2=(25)2+4(25)2+(18)225=340025=2345Area of triangle =12×|BC→|×|AD→ |=12×5×2345=34 [∵BC=5]

The vertices B and C of a triangle ABC lie on the line, x+23=y−10=z4 such that BC=5 units. Then the area (in sq. units) of this triangle, given that the point A(1,-1,2),is:

Let a point D on BC= (3λ,−2,1,4λ) AD→=(3λ−3)i∧+2j∧+(4λ−2)k∧∵ AD→ ⊥ BC→ , ∴AD→ . BC→=0⇒(3λ−3)+3+2(0)+(4λ−2)4=0⇒λ=1725Hence, D=(125,1,6825)|AD→|=(125−1)2+(2)2+(6825−2)2=(25)2+4(25)2+(18)225=340025=2345Area of triangle =12×|BC→|×|AD→ |=12×5×2345=34 [∵BC=5]

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The vertices $B$ and $C$ of a triangle $A B C$ lie on the line, $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $B C = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A (1, - 1, 2),$ is:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$2 \sqrt{34}$

$6$

$5 \sqrt{17}$

$\sqrt{34}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let a point D on BC= $(3 λ, - 2, 1, 4 λ)$

$\begin{array}{l} \vec{A D} = (3 λ - 3) \overset{\land}{i} + 2 \overset{\land}{j} + (4 λ - 2) \overset{\land}{k} \\ ∵ \vec{A D} ⊥ \vec{B C}, ∴ \vec{A D} . \vec{B C} = 0 \\ \Rightarrow (3 λ - 3) + 3 + 2 (0) + (4 λ - 2) 4 = 0 \\ \Rightarrow λ = \frac{17}{25} \\ H e n c e, D = (\frac{1}{25}, 1, \frac{68}{25}) \\ | \vec{A D} | = \sqrt{{(\frac{1}{25} - 1)}^{2} + {(2)}^{2} + {(\frac{68}{25} - 2)}^{2}} \\ = \sqrt{\frac{{(25)}^{2} + 4 {(25)}^{2} + {(18)}^{2}}{25}} = \sqrt{\frac{3400}{25}} = \frac{2 \sqrt{34}}{5} \\ A r e a o f t r i a n g l e = \frac{1}{2} \times | \vec{B C} | \times | \vec{A D} | \\ = \frac{1}{2} \times 5 \times \frac{2 \sqrt{34}}{5} = \sqrt{34} [∵ B C = 5] \end{array}$