Match the following Veginarray - hline -Column-I- -Column-IINhhline- -vectors 2 isqrt 3 i -2 sqrt 3 j - sqrt 3 k - and - 2 i - j -2 k - then the- end array A- a-rs- b-p- c-q- d-rB- a-prs- b-r- c-q- d-pC- a-s- b-p- c-q- d-prsD- a-q- b-pqs- c-r- d-s

The correct option is B a prs; b r; c q; d p(a) AB=2√(3i) 2√(3j)+√(3k) AC=2i+j 2k AB.AC=0 |AB|=3√(3) |AC|=3 ∠ A=90,∠ B=π/6,∠ C=π/3 (b) I ∫_0^π x sin^3 xdx ...

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Byju's Answer

Standard XII

Mathematics

Theorems for Continuity

Match the fol...

Question

Match the following
$\begin{matrix} C o l u m n - I & C o l u m n - I I (a) & T w o a d j a c e n t s i d e s o f a t r i a n g l e a r e r e p r e s e n t e d b y t h e & (p) & \frac{π}{2} v e c t o r s 2 \sqrt{3} i - 2 \sqrt{3} j + \sqrt{3} k a n d 2 i + j - 2 k t h e n t h e a n g l e s o f t h e t r i a n g l e a r e (b) & I f \int_{0}^{π} s i n^{3} x d x = 4 k, t h e n k = & (q) & \frac{2 π}{3} (c) & \int_{0}^{π} [2 s i n x] d x i s & (r) & \frac{π}{6} (d) & I f z_{1} a n d z_{2} s a t i s f y | z - 3 - 4 i | = 5 a n d | z_{3} - 3 - 4 i | = & (s) & \frac{π}{3} 5 \sqrt{2} t h e n A r g (\frac{z_{1} - z_{3}}{z_{2} - z_{3}}) i s \end{matrix}$

a-q; b-pqs; c-r; d-s

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a-rs; b-p; c-q; d-r

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a-prs; b-r; c-q; d-p

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a-s; b-p; c-q; d-prs

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Solution

The correct option is C a-prs; b-r; c-q; d-p
$(a) ¯ ¯¯¯¯¯¯ ¯ A B = 2 \sqrt{3 i} - 2 \sqrt{3 j} + \sqrt{3 k} ¯ ¯¯¯¯¯¯ ¯ A C = 2 i + j - 2 k ¯ ¯¯¯¯¯¯ ¯ A B . ¯ ¯¯¯¯¯¯ ¯ A C = 0 | ¯ ¯¯¯¯¯¯ ¯ A B | = 3 \sqrt{3} | ¯ ¯¯¯¯¯¯ ¯ A C | = 3 ∠ A = 90, ∠ B = \frac{π}{6}, ∠ C = \frac{π}{3}$

$(b) I - \int_{0}^{π} x s i n^{3} x d x 2 I = π \int_{0}^{π} x s i n^{3} x d x I = π \int_{0}^{\frac{π}{2}} s i n^{3} x d x = \frac{2 π}{3}$
$(c) \int_{0}^{π} [2 s i n x] d x = \int_{\frac{π}{6}}^{\frac{5 π}{6}} d x = \frac{2 π}{3}$
(d)One is the director circle of the other.

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