Show that the vector i ^ + j ^ + k ^ is equally inclined to the axes O X , O Y , and O Z .

A vector is equally inclined to the axes O X , O Y and O Z respectively, if its direction cosines are equal. The direction cosines of a vector a → = a 1 i ^ + a 2 j ^ + a 3 k ^ are a 1 a → , a 2 a → , a 3 a → . Let a → = i ^ + j ^ + k ^ Here, a 1 = 1 , a 2 = 1 , and a 3 = 1 Then, a → = 1 2 + 1 2 + 1 2 = 3 Therefore, the direction cosines of a → are 1 3 , 1 3 , 1 3 . Since the direction cosines are equal. Hence, the given vector is equally inclined to axes O X , O Y , and O Z .

Find the unit vector in the direction of the vector a → = i ^ + j ^ + 2 k ^ .

The unit vector in the direction of vector a → is a ^ = a → a → , if a → = a 1 i ^ + a 2 j ^ + a 3 k ^ then magnitude of a → is given by a → = a 1 2 + a 2 2 + a 3 2 To find: The unit vector in the direction of vector a → = i ^ + j ^ + 2 k ^ . a → = 1 2 + 1 2 + 2 2 = 1 + 1 + 4 = 6 ∴ a ^ = a → a → = i ^ + j ^ + 2 k ^ 6 = 1 6 i ^ + 1 6 j ^ + 2 6 k ^ Hence, the required unit vector is 1 6 i ^ + 1 6 j ^ + 2 6 k ^

Compute the magnitude of the following vectors: a → = i ^ + j ^ + k ^ ; b → = 2 i ^ - 7 j ^ - 3 k ^ ; c → = 1 3 i ^ + 1 3 j ^ - 1 3 k ^

The magnitude of any vector r → = x i ^ + y j ^ + z k ^ is given by r → = x i ^ + y j ^ + z k ^ = x 2 + y 2 + z 2 Given vectors are a → = i ^ + j ^ + k ^ ; b → = 2 i ^ - 7 j ^ - 3 k ^ ; c → = 1 3 i ^ + 1 3 j ^ - 1 3 k ^ Magnitude of a → is a → = 1 2 + 1 2 + 1 2 = 3 Hence, a → = 3 Magnitude of b → is b → = 2 2 + - 7 2 + - 3 2 = 4 + 49 + 9 = 62 Hence, b → = 62 Magnitude of c → is c → = 1 3 2 + 1 3 2 + - 1 3 2 = 1 3 + 1 3 + 1 3 = 1 Hence, c → = 1

Show that each of the given three vectors is a unit vector: 1 7 2 i ^ + 3 j ^ + 6 k ^ , 1 7 3 i ^ - 6 j ^ + 2 k ^ , 1 7 6 i ^ + 2 j ^ - 3 k ^ Also, show that they are mutually perpendicular to each other.

Let a → = 1 7 2 i ^ + 3 j ^ + 6 k ^ = 2 7 i ^ + 3 7 j ^ + 6 7 k ^ , b → = 1 7 3 i ^ - 6 j ^ + 2 k ^ = 3 7 i ^ - 6 7 j ^ + 2 7 k ^ c → = 1 7 6 i ^ + 2 j ^ - 3 k ^ = 6 7 i ^ + 2 7 j ^ - 3 7 k ^ We know that, Magnitude of a vector is the square root of the sum of the squares of its scalar components. ∴ a → = 2 7 2 + 3 7 2 + 6 7 2 = 4 49 + 9 49 + 36 49 = 1 b → = 3 7 2 + - 6 7 2 + 2 7 2 = 9 49 + 36 49 + 4 49 = 1 c → = 6 7 2 + 2 7 2 + - 3 7 2 = 36 49 + 4 49 + 9 49 = 1 . Since, magnitudes of each of the vectors, a → , b → and c → is 1 . Hence, a → , b → and c → are unit vectors. The dot product of two vectors is the sum of the product of their corresponding scalar components. ∴ a → ⋅ b → = 2 7 × 3 7 + 3 7 × - 6 7 + 6 7 × 2 7 = 6 49 - 18 49 + 12 49 = 0 b → ⋅ c → = 3 7 × 6 7 + - 6 7 × 2 7 + 2 7 × - 3 7 = 18 49 - 12 49 - 6 49 = 0 c → ⋅ a → = 6 7 × 2 7 + 2 7 × 3 7 + - 3 7 × 6 7 = 12 49 + 6 49 - 18 49 = 0 . Since, the dot product is 0 . Hence, the given three vectors are mutually perpendicular to each other.

Find the direction cosines of the vector i ^ + 2 j ^ + 3 k ^ .

Given vector, i ^ + 2 j ^ + 3 k ^ Direction cosines of a vector a → = a 1 i ^ + a 2 j ^ + a 3 k ^ are a 1 a → , a 2 a → , a 3 a → . Let a → = i ^ + 2 j ^ + 3 k ^ Here, a 1 = 1 , a 2 = 2 and a 3 = 3 ∴ a → = 1 2 + 2 2 + 3 2 = 1 + 4 + 9 = 14 Hence, the direction cosines of a → are 1 14 , 2 14 , 3 14 .

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

The given points are P 1 , 2 , 3 and Q 4 , 5 , 6 . ∴ P Q → = 4 - 1 i ^ + 5 - 2 j ^ + 6 - 3 k ^ = 3 i ^ + 3 j ^ + 3 k ^ P Q → = 3 2 + 3 2 + 3 2 = 9 + 9 + 9 = 27 = 3 3 We know that The unit vector in the direction of the given vector a → is a ^ = a → a → . Hence, the unit vector in the direction of P Q → is P Q → P Q → = 3 i ^ + 3 j ^ + 3 k ^ 3 3 = 1 3 i ^ + 1 3 j ^ + 1 3 k ^