If hat i + hat j , hat j + hat k , hat i + hat k are the position vectors of the vertices of a Δ A B C taken in order, then angle A is equal to

Question

If hat i + hat j , hat j + hat k , hat i + hat k are the position vectors of the vertices of a Δ A B C taken in order, then angle A is equal to (A) (π/2) (B) (π/5) (C) (π/6) (D) (π/4)

Tardigrade Admin · Accepted Answer

Let position vector of the vertices are

O A = \hat{i} + \hat{j}, OB = \hat{j} + \hat{k}

and

OC = \hat{i} + \hat{k}

Now,

A B = - \hat{i} + \hat{k}

and

A C = \hat{k} - \hat{j}

∴ cos θ = \frac{( A B ) \cdot ( A C )}{∣ A B ∣∣ A C ∣}

= \frac{( - i ^ + k ^ ) \cdot ( k ^ - j ^ )}{1 ^{2} + 1 ^{2} 1 ^{2} + 1 ^{2}}

= \frac{( k ^ ) ^{2}}{2 2} = \frac{1}{2}

\Rightarrow θ = \frac{π}{3}

Q. If $\hat{i} + \hat{j}, \hat{j} + \hat{k}, \hat{i} + \hat{k}$ are the position vectors of the vertices of a $Δ A BC$ taken in order, then $∠ A$ is equal to

$\frac{π}{2}$

$\frac{π}{5}$

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

Q. If i^+j^​,j^​+k^,i^+k^ are the position vectors of the vertices of a ΔABC taken in order, then ∠A is equal to

2π​

5π​

6π​

4π​

3π​