The two vectors A- and B- are drawn from a common point and C-A-B- then angle between A- and B- is

In Fig. 7.222, angle B is greater than 90 degrees and segement $A D ⊥ B C$ , show that
(i) $b^{2} = h^{2} + a^{2} + x^{2} - 2 a x$
(ii) $b^{2} = a^{2} + c^{2} - 2 a x$

Q. If

∣ ∣ ∣ ∣ ∣ \begin{matrix} a & a^{2} & 1 + a^{3} b & b^{2} & 1 + b^{3} c & c^{2} & 1 + c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

and the vectors

\to A = (1, a, a^{2});

\to B = (1, b, b^{2}); \to C = (1, c, c^{2})

are non-coplanar then the product

a b c =

Let $\to a = a_{1}^i + α_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k a n d \to c = c_{1}^i + c_{2}^j + c_{3}^k$ be three non-zero vectors such that $\to c$ is a unit vector perpendicular to both the vectors $\to a$ and $\to b$ . If the angle between $\to a$ and $\to b$ is $\frac{π}{6},$
then ${∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2}$ is equal to

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The two vectors →A and →B are drawn from a common point and →C=→A+→B then angle between →A and →B is

The correct option is B greater than 90∘ if C2<A2+B2 ∵→C=→A+→B∴C2=A2+B2+2ABcosθ If C2<A2+B2 then cosθ<0. Therefore θ>90∘

The two vectors $\to A$ and $\to B$ are drawn from a common point and $\to C = \to A + \to B$ then angle between $\to A$ and $\to B$ is

The correct option is B greater than $90^{\circ}$ if $C^{2} < A^{2} + B^{2}$

$∵ \to C = \to A + \to B ∴ C^{2} = A^{2} + B^{2} + 2 A B c o s θ$

If $C^{2} < A^{2} + B^{2}$ then $c o s θ < 0$ .

Therefore $θ > 90^{\circ}$