Find the unit vector in the direction of the vector P Q → , where P and Q are the points 1 , 2 , 3 and 4 , 5 , 6 respectively.

The coordinates of the points P and Q are given. So, the position vector of P and Q can be written as O P → = i ^ + 2 j ^ + 3 k ^ and O Q → = 4 i ^ + 5 j ^ + 6 k ^ , where O is the origin. Therefore, P Q → = O Q → - O P → = 4 i ^ + 5 j ^ + 6 k ^ - i ^ + 2 j ^ + 3 k ^ = 3 i ^ + 3 j ^ + 3 k ^ Unit vector in the direction of vector P Q → is given by P Q ^ = P Q → P Q → = 3 i ^ + 3 j ^ + 3 k ^ 3 2 + 3 2 + 3 2 = 3 i ^ + 3 j ^ + 3 k ^ 27 = i ^ + j ^ + k ^ 3 = i ^ 3 + j ^ 3 + k ^ 3

Find the unit vector in the direction of the vector P Q → , where P and Q are the points 1 , 2 , 3 and 4 , 5 , 6 respectively.

The coordinates of the points P and Q are given. So, the position vector of P and Q can be written as O P → = i ^ + 2 j ^ + 3 k ^ and O Q → = 4 i ^ + 5 j ^ + 6 k ^ , where O is the origin. Therefore, P Q → = O Q → - O P → = 4 i ^ + 5 j ^ + 6 k ^ - i ^ + 2 j ^ + 3 k ^ = 3 i ^ + 3 j ^ + 3 k ^ Unit vector in the direction of vector P Q → is given by P Q ^ = P Q → P Q → = 3 i ^ + 3 j ^ + 3 k ^ 3 2 + 3 2 + 3 2 = 3 i ^ + 3 j ^ + 3 k ^ 27 = i ^ + j ^ + k ^ 3 = i ^ 3 + j ^ 3 + k ^ 3

If P ≡ 1 , 5 , 4 and Q ≡ 4 , 1 , - 2 , find the direction ratios of P Q → .

The coordinates of points P and Q are given, so their position vectors can be found out as O P → = i ^ + 5 j ^ + 4 k ^ and O Q → = 4 i ^ + j ^ - 2 k ^ , where O is the origin. P Q → = 4 i ^ + j ^ - 2 k ^ - i ^ + 5 j ^ + 4 k ^ = 3 i ^ - 4 j ^ - 6 k ^ Direction ratios of the vector P Q → are the coefficients of i ^ , j ^ and k ^ , i.e., 3 , - 4 , - 6 respectively.

If a → = i ^ + j ^ + k ^ , b → = 2 i ^ - j ^ + 3 k ^ and c → = i ^ - 2 j ^ + k ^ , find a unit vector parallel to the vector 2 a → - b → + 3 c → .

In the question, vectors a → = i ^ + j ^ + k ^ , b → = 2 i ^ - j ^ + 3 k ^ and c → = i ^ - 2 j ^ + k ^ are given. By substituting these in 2 a → - b → + 3 c → , the vector can be obtained. Let us say that the vector so obtained is A → . Therefore, A → = 2 a → - b → + 3 c → = 2 i ^ + j ^ + k ^ - 2 i ^ - j ^ + 3 k ^ + 3 i ^ - 2 j ^ + k ^ = 2 - 2 + 3 i ^ + 2 + 1 - 6 j ^ + 2 - 3 + 3 k ^ = 3 i ^ - 3 j ^ + 2 k ^ The unit vector of A → = 3 i ^ - 3 j ^ + 2 k ^ is obtained as u A → = A → A → . Therefore, u A → = 3 i ^ - 3 j ^ + 2 k ^ 3 2 + - 3 2 + 2 2 = 3 i ^ - 3 j ^ + 2 k ^ 9 + 9 + 4 = 3 i ^ - 3 j ^ + 2 k ^ 22 . In the question, unit vector parallel to A → is asked, i.e., the parallel vector can be in the direction as well as opposite to the direction of the vector A → . So, the answer would be u A → = ± 1 22 3 i ^ - 3 j ^ + 2 k ^

If a → = i ^ + 2 j ^ - k ^ and b → = 3 i ^ + j ^ - 5 k ^ , find a unit vector in the direction of a → - b → .

Two vectors are given as a → = i ^ + 2 j ^ - k ^ and b → = 3 i ^ + j ^ - 5 k ^ , their difference is obtained as a → - b → = i ^ + 2 j ^ - k ^ - 3 i ^ + j ^ - 5 k ^ = 1 - 3 i ^ + 2 - 1 j ^ + - 1 + 5 k ^ = - 2 i ^ + j ^ + 4 k ^ To find a unit vector in the direction of a → - b → , u → = a → - b → a → - b → u → = - 2 i ^ + j ^ + 4 k ^ - 2 2 + 1 2 + 4 2 = - 2 i ^ + j ^ + 4 k ^ 4 + 1 + 16 = - 2 i ^ + j ^ + 4 k ^ 21 = - 2 i ^ 21 + j ^ 21 + 4 k ^ 21

If the position vectors of P → , Q → , R → and S → be respectively, 2 i ^ + 4 k ^ , 5 i ^ + 3 3 j ^ + 4 k ^ , - 2 3 j ^ + k ^ and 2 i ^ + k ^ , prove that R S → is parallel to P Q → and is two-third of P Q → .

Let, O be the origin. From the question, the position vectors of each point can be written as: O P → = 2 i ^ + 4 k ^ , O Q → = 5 i ^ + 3 3 j ^ + 4 k , ^ O R → = - 2 3 j ^ + k ^ , O S → = 2 i ^ + k ^ Using the above vectors, R S → and P Q → can be formulated as: R S → = ( 2 i ^ + k ^ ) - ( - 2 3 j ^ + k ^ ) = 2 i ^ + 2 3 j ^ = 2 ( i ^ + 3 j ^ ) P Q → = ( 5 i ^ + 3 3 j ^ + 4 k ^ ) - ( 2 i ^ + 4 k ^ ) = ( 5 - 2 ) i ^ + ( 3 3 + 0 ) j ^ + ( 4 - 4 ) k ^ = 3 i ^ + 3 3 j ^ = 3 ( i ^ + 3 j ^ ) Comparing both the vectors above, R S → = 2 ( i ^ + 3 j ^ ) , P Q → = 3 ( i ^ + 3 j ^ ) ⇒ R S → = 2 3 P Q → For two vectors to be parallel, they must satisfy the condition a → = m b → , where m is a scalar. So, from the relation P Q → , R S → are parallel.

Prove that the vectors 2 i ^ - j ^ + k ^ , i ^ - 3 j ^ - 5 k ^ and 3 i ^ - 4 j ^ - 4 k ^ form a right-angled triangle.

Let A B C be the triangle. Then consider A → = 2 i ^ - j ^ + k ^ , B → = i ^ - 3 j ^ - 5 k ^ and C → = 3 i ^ - 4 j ^ - 4 k ^ . The sides ` A B , B C & C A of the triangle can be found using position vectors. A B → = i ^ - 3 j ^ - 5 k ^ - 2 i ^ - j ^ + k ^ = - i ^ - 2 j ^ - 6 k ^ B C → = 3 i ^ - 4 j ^ - 4 k ^ - i ^ - 3 j ^ - 5 k ^ = 2 i ^ - j ^ + k ^ C A → = 2 i ^ - j ^ + k ^ - 3 i ^ - 4 j ^ - 4 k ^ = - i ^ + 3 j ^ + 5 k ^ The length of the sides of the triangle can be obtained by computing the magnitude of the above vectors. A B → = ( - 1 ) 2 + ( - 2 ) 2 + ( - 6 ) 2 = 1 + 4 + 36 = 41 B C → = 2 2 + ( - 1 ) 2 + 1 2 = 4 + 1 + 1 = 6 C A → = ( - 1 ) 2 + 3 2 + 5 2 = 1 + 9 + 25 = 35 In order to check if it is a right-angled triangle, we have to apply Pythagoras theorem. Side A B has the highest magnitude and is taken as the hypotenuse. B C → 2 + C A → 2 = A B → 2 ⇒ 6 2 + 35 2 = 41 2 ⇒ 6 + 35 = 41 ⇒ 41 = 41 Since the Pythagoras theorem is satisfied, the triangle A B C is a right-angled triangle.

If the position vectors of P , Q , R and S be respectively 2 i ^ + 4 k ^ , 5 i ^ + 4 j ^ + 4 k ^ , - 4 i ^ - 8 j ^ + k ^ and 2 i ^ + k ^ , prove that R S is parallel to P Q and is twice of P Q .

Since the position vectors are given in the question, the vectors can be computed as R S → = ( 2 i ^ + k ^ ) - ( - 4 i ^ - 8 j ^ + k ^ ) = 6 i ^ + 8 j ^ = 2 ( 3 i ^ + 4 j ^ ) P Q → = ( 5 i ^ + 4 j ^ + 4 k ^ ) - ( 2 i ^ + 4 k ^ ) = 3 i ^ + 4 j ^ From the above equations, it is clear that R S → = 2 P Q → . For two vectors to be parallel, they must be related by means of a scalar m , say, a → = m b → . Here, 2 is the scalar that relates vectors R S → and P Q → and hence they are parallel.

The position vectors of the points P , Q , R and S are i ^ + j ^ + k ^ , 2 i ^ + 5 j ^ , 3 i ^ + 2 j ^ - 3 k ^ and i ^ - 6 j ^ - k ^ . Prove that the lines P Q and R S are parallel and find the ratio of their lengths.

Using the position vectors given in the question, the vectors can be computed as R S → = i ^ - 6 j ^ - k ^ - 3 i ^ + 2 j ^ - 3 k ^ = - 2 i ^ - 8 j ^ + 2 k ^ = - 2 i ^ + 4 j ^ - k ^ P Q → = 2 i ^ + 5 j ^ - i ^ + j ^ + k ^ = i ^ + 4 j ^ - k ^ From the above equations, R S → = - 2 P Q → . The ratio of the length of vectors is R S → P Q → = 1 2 ⇒ R S → : P Q → = 1 : 2 For these vectors to be parallel, they must be related by a scalar m , say, a → = m b → , where m can be positive or negative. Here, R S → = - 2 P Q → , where - 2 is the scalar, indicating that the two vectors are parallel to each other. They have the same magnitude but are opposite in direction.